Wlementary M.echantcs 




Class _^. 
Book_ 



A Q^ r\r^f 



Copyright N^. 



COFH?IGHT DEPOSm 



Elementary Mechanics 



Elementary Mechanics 



PREPARED BY THE 

DEPARTMENT OF MATHEMATICS 

OF THE UNITED STATES 

NAVAL ACADEMY 



ANNAPOLIS, MD. 

THE UNITED STATES NAVAL INSTITUTE 

1920 






Copyright, 1920, by 

J. W. CONROY 

Trustee for U. S. Naval Institute 
Annapolis, Md. 



SEP 27 1£20 



BALTIMORE, MD., U. S. A. 



©CU576574 



PEEFACE 

This text has been prepared for a short course in mechanics. 
The attempt has been made to present the fundamental prin- 
ciples of mechanics in a simple manner and to give a rigorous dis- 
cussion of each topic to the extent that it is treated. 

The text has been written by the following members of the 
Department of Mathematics of the U. S. Naval Academy: 
Messrs. W. J. King, J. B. Eppes, J. A. Bullard, John Tyler, 
Arthur Kiernan, J. N. Galloway, Alex. Dillingham, G. R. 
Clements, H. M. Eobert, Jr., L. S. Dederick, L. T. Wilson, W. F. 
Shenton, and R. P. Johnson. Other members of the department 
have offered suggestions and contributed problems. 

The work of editing has been done by Commander H. L. Rice, 
(Math.), U. S. N., and Messrs. Bullard and Dederick. 



CONTENTS 

CHAPTER I 

Vectors page 

1. Vector and Scalar Quantities 17 

2. Graphical Representation of Vector Quantities 17 

3. Equal Vectors 18 

4. Multiplication of a Vector Quantity by a Scalar Quantity. . . 18 

5. Addition of Two Vectors 18 

6. Subtraction of Vectors 20 

7. Addition of Several Vectors 20 

8. Relations Between the Magnitudes of Two Vectors and 

Their Resultant 22 

9. Resolution of Vectors 22 

10. Illustrative Example 24 

Problems 25 

CHAPTER II 
Statics of a Particle 

11. Statics 27 

12. Force 27 

13. Units of Force 27 

14. Forces as Vectors: Parallelogram of Forces 28 

15. Polygon of System of Concurrent Forces 28 

16. Equilibrium 30 

17. Illustrative Examples 30 

Problems 32 

18. Types of Forces 34 

19. Action and Reaction 34 

20. Law of Transmissibility ' 35 

21. Illustrative Examples 36 

Problems 39 

22. Resolution of Forces 41 

23. Algebraic Method for Finding the Resultant of a System cf 

Concurrent Forces 42 

24. Algebraic Conditions for Equilibrium 44 

Problems 47 



3 Contents 

V\V,K 

25. Friction 40 

2G. The Angle of Repose 50 

27. Illustrative Problems on Friction 51 

Problems 63 

Keviow Problems 55 



CllAPTFK 111 

FoiU'KS AeriM) on .v Uu;in I'opy 

2S. Definition 61 

29. INlotion of a Rigid Body 61 

30. Forces Acting on a Rigid Podv 62 

31. The Moment of a Force About a Pino 62 

32. The Moment of a Force About a Point 63 

33. Sign of a Moment 64 

34. Theorem: The moment about a point () of a force /■' acting 

at a point .4 is equal to 0.4 multiplied by the component 

of F In a direction perpendicular to OA •. i)4 

35. Theorem: The sum of the moments about any point of 

any number of concurrent forces is equal to the moment 

about of their resultant 65 

36. Non-parallel Forces 65 

37. Parallel Forces (U; 

3S. Cases I and II 66 

39. The Moment of the Resultant of Two Parallel Forces (Ex- 

cept Case Iin 67 

40. The Resultant of Any Number of Parallel Forces 68 

41. Coordinates of the Center of .\ny Number of Parallel Forces 6S 
Problems OS) 

42. Case III. Couples 71 

43. Moment of a Couple 71 

44. Composition of Couples 7ii 

45. A Couple and a Force Replaced by a Single Force 73 

46. Theorem: Any number of eoplanar forces acting on a rigid 

body is equal to a force acting at any desired point in the 

plane and a couple 74 

Problems 75 

Review Problems 76 



(Jo NT K NTS 9 

CHAI'TER IV 
Statich ok a Rioh) Hody paok 

M. [<:(|ullil)riurn of a KiKid Hody 7H 

4K. If a ri^id l)0(Jy is in (;(iuilil)riurn undor throo forces, the throe 

forces pass tlirou^h a point or are parallel 78 

41). Illustrative Examples , 70 

i'robhims HO 

50. Two-force I'ieces HI 

r>l. Method of Sections HI 

Problems H4 

52. Maxwell's Method ^^> 

f'rohlenis H8 

5:5. ''rhr(!e-force Pieces 00 

i'rohlems 02 

Review Problems 03 



(MlM'^riOR V 

ClONTKIt OK GltAVITY AND MoMKNT OK iNKItTIA 

54. Gravitational l^'orces 08 

55. ( 'enter of Gravity 08 

5(). Goordi nates of ('entc^r of (Jravity 08 

57. Application of the Galculus 101 

58. Mass Goneentrated at Genter of (Gravity 102 

59. Genter of Mass; Gentroid 102 

60. Illustrative Examples 102 

Problems 1 07 

Gl. Center of Gravity of Gomposite l?odies 108 

Problems 1 10 

G2. Theorems of Pappus and (Juldinus HI 

Problems ; Ill} 

iVA. yvpproximate Integration ll.'J 

(!4. Simpson's Rule llf) 

()5. Illustrative ICxa'mph.^s 1 17 

Problems 118 

00. Moment of Inertia 1 20 

G7. Illustrative Examples 120 

Problems 123 



10 Contents 



PAGE 

68. Theorem of Parallel Axes 123 

69. Radius of Gyration 124 

70. Illustrative Examples f . 125 

Problems* .^ 127 

71. Polar Moment of Inertia 128 

72. Illustrative Examples 129 

Problems 131 

CHAPTER VI 
Fluid Pressure 

73. Properties of Fluids 132 

74. Pressure 133 

75. Theorem: The pressure at any point of a fluid is the same 

in all directions 134 

76. Pressure Varies with Depth 135 

77. Transmissibility of Pressure 136 

78. Total Pressure Force 137 

79. Cylindrical Surfaces 139 

Problems 141 

80. Center of Pressure 142 

81. Examples 143 

82. General Illustrative Examples 145 

Problems 148 

Review Problems 149 

CHAPTER VII 
Deformable Bodies 

83. Introduction ' 152 

84. Normal Stress 153 

Problems 154 

85. Longitudinal Strain 154 

86. Stress-Strain Diagram 155 

87. Hooke's Law and the Modulus of Elasticity 155 

Problems 156 

88. Working Stress and Factor of Safety 156 

Problems 157 

89. Shear 157 

Problems 158 



COXTENTS 11 

\ 

PAGE 

90. Equal Strength 159 

Problems 160 

91. Beams 160 

92. Kinds of Stresses in Beams 162 

93. Shearing Force 163 

94. Algebraic Sign of the Shearing Force 164 

95. Shearing Force Diagrams 164 

Problems 166 

96. Bending Moment 167 

97. Sign of the Bending Moment 169 

98. Bending Moment Diagrams 170 

Problems 172 

99. Relations Connecting Loading, Shearing Force and Bending 

Moment 173 

100. Illustrative Examples 175 

101. Dangerous Section 178 

Problems 180 

102. Distribution of Stress 180 

103. Bending 181 

104. Location of Neutral Axis 182 

105. Moment of Internal Couple 183 

106. Units of Measurement 184 

Problems 185 

107. Section Modulus 186 

Problems 186 

108. Superposition 187 

Problems 187 

109. Torsion 188 

Problems 190 

Review Problems 190 

CHAPTER VIII 
Rectilinear Motion 

110. Velocity 194 

Problems 195 

111. Acceleration 195 

Problems 196 

112. Equation ads — v.dv 197 

Problems 197 



12 Contents 

PAGE 

113. Relative Velocity , „ . . 198 

Problems 198 

114. Mass and Momentum 199 

115. Newton's Laws of Motion 200 

116. Conservation of Momentum 202 

Problems 203 

117. Elastic Impact 204 

Problems 206 

118. The Equation of Motion 206 

119. Units of Mass and Force 207 

Problems 207 

120. Falling Bodies 208 

Problems 208 

121. Absolute Units 209 

122. Engineer's Units 209 

123. Summary of Systems of Units 210 

Problems . . . .• 211 

124. Constant Force 212 

Problems 213 

125. The Relation F/F' —a/ a' 213 

Problems 214 

126. Relative Acceleration 214 

Problems 215 

127. Motion on an Inclined Plane. 215 

Problems 218 

128. Atwood's Machine 218 

Problems 220 

129. Other Problems on Connected Bodies 220 

Problems 223 

130. Motion Under a Variable Force. 223 

131. Force Depending on the Time 224 

132. Force Depending on the Distance 226 

133. Simple Harmonic Motion 228 

Problems 229 

134. Force Depending on the Velocity 230 

Problems 230 

Review Problems 230 



Contents 13 

CHAPTER IX 
Work and Energy page 

135. Work 232 

136. Work Done by a Constant Force 232 

Problems 234 

137. Work Done when the direction of the displacement makes 

an angle e with line of action of the force 234 

Problems 236 

138. Energy 237 

139. Expression for Kinetic Enerry 238 

Problems 238 

140. Work Done by Force Constant in Direction but Variable in 

Magnitude 239 

141. Work Done When Variable Force Makes Angle d with Direc- 

tion of Displacement 242 

Problems 242 

142. Power 243 

Problems 244 

143. Relation Between Kinetic Energy and Work 245 

Problems 247 

Review Problems 248 



CHAPTER X 
Curvilinear Motion 

144. Vector Velocity 251 

145. Vector Acceleration 252 

146. Composition and Resolution of Velocities and Accelerations 252 
Problems 255 

147. Velocity and Acceleration Independent of Origin 256 

148. Tangential and Normal Acceleration 257 

Problems 258 

149. Circular Motion 259 

150. Units 259 

Problems 260 

151. Linear and Angular Displacement, Velocity, and Acceleration 260 
Problems 262 

152. Newton's Laws and the Differential Equation of Motion 262 



14 Contents 



PAGE 

153. Components of Force and Acceleration 263 

154. Free Motion and Constrained Motion 264 

155. Motion of a Projectile 265 

Problems 267 

156. Centripetal and Centrifugal Force 268 

Problems 270 

157. Conical Pendulum 270 

Problems 272 

158. Surface of a Rotating Fluid 273 

Problems 275 

159. Simple Pendulum 275 

160. The Seconds Pendulum 277 

161. Cycloidal Pendulum 277 

Problems 279 

162. Motion Along a Smooth Curve in a Vertical Plane 279 

Problems 281 

163. Work 281 

164. Kinetic Energy 282 

Problems 283 



CHAPTER XI 

Dynamics of a Rigid Body 

165. Rigid Body 286 

166. Plane Motion 286 

167. External and Internal Forces : 286 

168. Translation 287 

169. Motion of the Center of Gravity in Translation 288 

170. Examples 288 

Problems 289 

171. Rotation 290 

172. Equation of Angular Motion 290 

173. Examples 292 

Problems 293 

174. The Compound Pendulum 294 

175. Centers of Suspension and Oscillation 294 

Problems 295 

176. The Torsion Balance 296 

Problems 298 



Contexts 15 

PAGE 

177. Kinetic Energy of Rotation and the Work-Energy Relation 298 
Problems 300 

178. Power Transmitted by Shafts 301 

Problems 302 

179. Translation and Rotation; Plane Motion 302 

180. Examples 305 

Problems 306 

Review Problems 306 

Answers 310 



ELEMENTARY MECHANICS 



CHAPTER I 
VECTORS 

1. Vector and Scalar Quantities. — The quantities with which 
we shall be concerned in the study of Mechanics may be divided 
into two general classes, namely, those which involve both magni- 
tude and direction and those which involve magnitude only. The 
former are called vector quantities and the latter scalar quantities. 

Scalar quantities are of a type with which we are all familiar, 
though we may not have designated them by this term ; they are 
quantities which can be measured by means of a scale, hence the 
Qame. Temperature, time, density, voltage, mass, and many 
Dther quantities of every-day experience are illustrations of quan- 
tities which can be measured on one type of scale or another. 

But not all quantities can be so measured. Thus, as will be 
shown as each is treated, force, velocity and acceleration re- 
quire not only magnitude but direction for their determination. 
rhese are examples of vector quantities. 

2. Graphical Representation of Vector Quantities. — A vector 
quantity is represented by the segment 

of a straight line with an arrow at one 
end to indicate the sense. In Fig. 1 
the vector a is represented by the line 
AB from the initial point A to the 
terminus B. The arrow-head indi- A ' 
3ates that the direction along the line Pj-q_ i 

is from A to B. The length AB, or a, 

represents the magnitude of the vector. The student must 
clearly bear in mind that AB is not the same vector as BA. 
2 




18 Elementary Mechanics 

3. Equal Vectors. — Two vectors are smd to he equal when they 

have the same magnitude and direc- 
tion. From this it follows readily 
that we can move a vector at pleasure 
so long as we do not alter its magni- 
tude or change its direction. In 
Fig. 2 the vectors a and ^ are drawn 
equal in length and parallel in direc- 
FiG. 2. tion; hence we may write a = /?. On 

the other hand, we say y = — a be- 
cause the direction of y is opposite to that of a, although their 
lengths, and hence their magnitudes, are the same. 

4. Multiplication of a Vector Quantity by a Scalar Quantity. — 
The product of a vector quantity 

and a scalar quantity is a vector ^^^.. — ^* 
quantity. The resultant vector is ^ 

one whose direction is the same 
as that of the given vector and 
whose magnitude is the magni- 
tude of the given vector multi- pia. 3. 
plied by the scalar. 

If the scalar is negative the sense of the vector direction is 
reversed. 

Thus, if a is the vector which represents the given quantity, 
|a would be represented by /3 while —fa would be represented 
by y. 

5. Addition of Two Vectors. — Move one of the vectors until 
its origin falls upon the terminus of the other vector; complete 
the triangle by drawing a vector, the origin of which coincides 
with that of the first vector. This vector is defined as the sum of 
the two given vectors and is called the resultant. In Fig. 4, 
given any two vectors as a and j8, move ^ to the terminus 5 of a 



Vectors 



19 



"We have then the 




oT 



and draw AC with its sense from A to C 
veetor equation 

y = « + /?. 

We eonhi secure this same re- 
sultant bv what is called tlie 
paralhdograni method, as fol- 
lows : 

Move one of tlie vectors until . 

its origin coincides with that oL' 
the other vector, compleie the 
paraHelog-ram o\' whirli these 
two vectors form adjacent ^^ 

sides, and then draw a vector p,*— 

which has the common ori^'in ol' 
the given vectors for its origin 
and which is a diagonal of the parallelogram. As an illustra- 
tion of the paralUdogram method consider Fig. 5, in which the 
two vectors a and /S of Fig. -i are again used. The origin of 
veetor /3 is put at the origin of vector a and the parallelogram 



Y 



c< 




Fig. 4. 



D . 




Fig. 



/ 

,/V 



/ 
/ 



A'B'C'D' is completed. Draw the vector A'C. This is the resul- 
tant of the two vectors a and /3 by the parallelogram method. The 
vector (3' is equal to the vector l3. Hence the vector triangle 



20 Elementary Mechanics 

A'B'C is the same as the vector triangle of Fig. 4 and A'C is 
equal to y. 

Further^ since a is at the terminus of jS we have the vector 
equation 

/S + a' = y; similarly a + /?' = y 

But, a = a' and /3 — (3' and it follows at once that whether a is 
applied at the terminus of /3 or vice versa the result is the same. 
Consequently. the addition of the vectors is independent of the 
order in which they are taken. 

6. Subtraction of Vectors. — Subtraction in algebra is equiva- 

lent to the addition of a negative 

/ quantity; likewise to subtract 

/P vectors lue reverse the sense of 

oC / (\-.-(K ^^^^ ^^^ to he subtracted and add 

it to the other. In Fig. 6 the 

D ~;^ '^ difference a — ^ of the two vectors 

/ \ Y ^'^ / is represented by 8. The sum 

/ ^\ /n a + /? is represented by y. If, 

/ ^^ M / however, we draw the other diag- 

N^ qC / onal of the parallelogram, ABCD, 

^\^ / , we can readily see from geometry 

^ ^>^ / I that DB is equal and parallel to 

^r^ AE. From this it follows at once 

Fig. 6 . that the sum and difference of 

two vectors form the diagonals of the parallelogram deter- 
mined by them. 

7. The Addition of Several Vectors consists merely of the 
successive application of the triangle law for two vectors. Given 
any system of vectors; in order to find their sum we choose any 
two of the vectors and secure their resultant. From the above 
this resultant is unique and is a vector ; consequently, when this 



Vectors 21 

resultant is added to any other vector of the system the new 
resultant is unique. This process is carried on until every vector 
of the system has been included in the addition. 



Illustration. — To find the sum of the four vectors a, /?, y, 8: 
We here apply ^ to the terminus of a and by the triangle method 
we secure the resultant vector a + 13; to the terminus of this we 
apply y and secure a + ^ + y; to the terminus of the latter we 
apply 8 and get the resultant of the four vectors, a + ^ + y + 8. 







• c 






/"f 






• / 






/ / 






• / 


i . 




/ 


• 




A 


y 




/ 


/ 




/ 


y , 




/ 


.y\ 




Ja... 


*• oC 




^B 



Fig. 



The vectors and the resultant together form a polygon. As 
a matter of practise it is not necessary to draw the diagonals of 



22 



ELE:iIEXTAKY MeCHAXICS 



the polygon ; merely apply (3 to the terminus of a, y to that of /?, 
and so forth; complete the polygon by drawing the vector from 
the starting point to the terminns of the last vector applied. 

8, Relations Between the lEagnitiides of Two Vectors and 
Their Eesnltant. — Let a and p, Mg. S, be two vectors and y 
their resnltant. Let be the angle between the directions of a 
and y and ^ be the angle between the direction of a and that of /3. 
We will denote the lengths of a, jS, and y by a, b, and c, respect- 
ively. Then from trigonometry we have 
c- = fl- + Z) - + 2a6 . cos cf> 



and tan 



h . sin 6 



a + b ' cos 

9. Kesolntion of Vectors. — The projection of a vector npon 
a line is called the component or the resolved part of the vector 
Y 




-.w 

/;?>. 


1 ^ . 




1 V 

1 / o<^ 



Fig. 9. 



Fig. 10. 



along that line. For example, in Fig. 9, a^ and ay are the pro- 
jections of a along the .r- and ^-axes respectively, and are called 
the X- and ^-components of a. 

flj.=fl. cos 0, 

Qy — a sin B, 



Vectors 23 

There is a distinction between the component and resolved part 
of a vector which it is important to understand clearly. A re- 
solved pai:t is always a component but the reverse is not the case. 
For example, let us draw the axes O'm and O'n through 0' , 
making O'm parallel to OX but O'n oblique to OY. Then the 
components of a along these axes are am and an respectively. 
However, these are not resolved parts. The resolved part of a 
along O'm is ax just as before (since O'm is parallel to OX) 
while the resolved part of a along O'n is the projection of a 
along that line. 

In br-ief, resolved part is the projection of the vector along a 
line and it is not equal to the component along that line unless 
the angle between the components is a right angle. 

In three dimensions the projections are illustrated in Fig. 10, 
where the components along the three axes are ax, ay, and a^. 
From this we secure the vector equation 

a = ax-\-ay-\-as 

and the projections or components : 

ax — aco^6^, 
tty — acos Oof 
az = a cos ^3 . 

When ^3 = 90° t4ien 0^ = 6 and (9o = 90-^ and we have the case 
of the plane ; as in Fig. 9, 

ax — a cos 6, 

ay = a sin 6. 

If we have several vectors in space and denote by px, py and pz the 
sum of the components along the x-, y-, and ;z-axes respectively 
we have 



r—yrx'^-\-i\/ + rz^, 
where r equals the length of the resultant p. The angles which p 



24 



Elementary Mechanics 



makes with the three coordinate axes are determined by the 
equations 



cos 0-^ 



cos^o = ^. 



cos t/, = 



where 0-^, 6^, and 6^ represent the angles which p makes with the 
X-, y-, and ;z-axes respectively. 

In case all the vectors lie in one plane, say the xy plane, we 
have r« = 0, ^3 = 90% 6^:^:0, 0^ = 90° -0 and 



r=Vra;^ + ry^, 



cos 



sm 



or tan^=-^. 



10. 



Illustrative Example 



An automobile travels 10 miles E 30° S, then 20 miles W, then 
30 miles N 45° E, then 5 miles N 30° W. Find its final position. 




Vectors 25 

r^ = 10 cos 30° + 20 cos 180° + 30 cos 45° + 5 cos 120° 

= 5V3-20 + 15V2-2.5 

= 7.37 miles • 
r^= - 10 cos 60° +^0 cos 90° + 30 cos 45° + 5 cos 30° 

= -5 + + 15V2 + 2.5V3 

= 20 .54 miles; 
r=V7.37- + 20.542 = 21.8 miles; • 

ianO= -^ = ?^ ^ = 70°16'. 



7.37 



Problems 



1. Take any four vectors and draw all the vector polygons 
possible and show graphically that all the polygons give the same 
resultant vector. 

2. Find the resultant vector of the two vectors whose magni- 
tudes are 5 and 3 respectively and which make an angle of 60° 
with one another. 

3. Show that the complex numbers 3 + 4i and 2 + 3i can be 
represented as vectors acting from a common point, which is the 
origin of real and imaginary numbers. Show that their sum 
obeys the parallelogram law. 

4. Show that the sum of the vectors represented by the 
medians of any triangle, all drawn from the vertices to the 
opposite sides, is zero. 

5. ABCD is a parallelogram and Y the intersection of the diag- 
onals, is any point within or without the plane of the paral- 
lelogram. Show that the resultant of the vectors OA^ OB, OC, 
OD is 4 OY. 

6. Vectors are represented in magnitude and direction by lines 
drawn from any point to the 8 corners of a cube. Show that 
their sum is represented by 8 OF when Y is the center of the cube. 

7. A man walks 4 miles north, then 6 miles northeast, then 3 
miles east, then 8 miles south. Determine by diagram his distance 
and direction from the starting point. 

8. If ABCD is a quadrilateral, prove that the sum of the vec- 
tors AB, AD, CB, and CD is represented by four times the line 
joining the middle points of the diagonals. 



26 Elementary Mechanics 

9. The magnitudes and direction angles of certain vectors in 
space are given below. Find the resultant in magnitude and 
direction in each case. 

(a) 10, ^TT, ^TT, Itt; 5, in, i-rr, Itt. 

(b) 6, Itt, Itt, Itt; 4, ^tt, -^tt, Itt. 

*10. Giventhevectorsa (3, -2),6(5,0),c(-10,6),^(7,7). 
Construct the figures and find the resultant of the following 

(a) a + 2b-3c; 

(b) 2a-h + c + 2d; 

(c) 3a + 4c-tZ; 

(d) 2a-3b-2c-2cU 

11. A point undergoes three displacements of 1, 2, and 3 units, 
respectively, in directions parallel to the sides of an equilateral 
triangle taken in order. What is the resultant displacement? 

* 12. Find the resultant of three coplanar vectors a, h, c, whose 
components are (3, —2), (2, 6), (—7, —1), respectively. 

* 13. Given in the xy plane a vector whose axial components 
are ( — 2, 1 ) . Find its resolved part along the directed line from 
the origin to the point (2, 1). 

14. A vector drawn east has a length of 20 inches, and one 
drawn northeast a length of 30 inches. What is their vector sum ? 

* 15. Find the resultant of three coplanar vectors a, &, c, whose 
components are (4, —1), (—3,8), ( — 7, —5) respectively. 

* (a, h) denotes a vector drawn from the origin to the point 

(a, &). 



Statics of a Particle 27 



CHAPTER II 
STATICS OF A PARTICLE 

11. Statics is that portion of Mechanics which treats of bodies 
at rest. The treatment of bodies which are in uniform motion 
is closely allied to the treatment of bodies at rest^ but we will 
restrict ourselves in the present chapter to bodies which are 
actually at rest. 

In Mechanics the word particle is used to designate a portion 
of matter whose dimensions are negligible. Practically, how- 
ever, a great many bodies whose dimensions are not negligible 
can be treated as if this were true. The body is considered as 
concentrated at a point and hence treated as if it were a particle. 

12. Force. — When the action of two hodies upon each other 
tends to produce a change in the motion of the bodies^ we say 
that the bodies are exerting force upon each other. Thus, if a 
man pushes against a solid wall he tends to move the wall in the 
direction in which he is pushing. Force is then the tendency to 
produce motion and we recognize it as a push or a pull. It has 
magnitude, line of action, sense and point of application. All 
forces are distributed, that is, act over an appreciable area or 
volume, but frequently this is relatively small and can be re- 
garded as a point. Forces are considered then as concentrated 
forces acting at a particular point. 

13. Units of Force. — The most common example of force is 
that due to the attraction of the earth upon a body. This force 
is called weight. It varies slightly for different latitudes and 
elevations and would vary considerably at great distances from 
the surface of the earth. Ordinarily no account need be taken 
of this variation at different places, for errors introduced into 



28 



Elementary Mecii an ics 



Y, 



oiiii'ineoriiig (.'alculaiion, bocauso o( it, aro }n'aciically alwav?^ 
negligiblo. 

The unit o\' (ovcc is tlio uoi^ht o\' a givon quantity of matter 
as specified by law. This unit is the pouuil in most English 
speakino- eonntries and the kiloi^rani elsowhoro. The kilogram 
is 'J/vO-U)l pounds avoii\lu[KMs. Innh units will bo used in this 
text. 

AVhenever the ton is mentioned it is to be nnderstood, unless 
otiierwise specitied, that ^ve mean the so-ealled "long ton " which 
is '2240 pounds. 

14. Forces as Vectors; ParallelogTam of Forces. — A force 

\vhieh has a given point of appli- 
cation is fully determined by its 
nuigtiititde and direction. Let us 
take two forces A and /> aet- 

*^ ing at a eoinnum point: draw a 

vector a whose direction is the 
*^ same as that o( A and whose 

^'u-- 1-. length represents the magnitude 

of .4. Draw also a vector ,S in the same manner for B. Let y 
represent the residtant a-r,o ol' these two vectors. Ir has been 
Tcrilied by experiineni that y represents in lyagnitnde and direc- 
tion that force whii.h has the same ctl'ec-t as the two forces .4 and 
B acting siinidtancou<ly : in other words, ihe resuliant y repre- 
sents ihc rcsullanl of A anJ />. \Vc sluill fhcrefore assume a 
paralleloijrani law of forces corrc<po)idiiuj to the parallelogram 
met 110 J for rectors. 

15. Polyg-on of System of Concurrent Forces. — Forces Avlvicli 
have a conimon point o( ap{>lication are called concurrent forces. 
If a/u/ numlh'r of forces are conciirreni the resultant of the 
si/stem is a force irhich is represented hi/ the resultant vector 
obtained /'// considering the forces as rectors and drawing the 




Statics of a Particle 



29 



vector polygon. Thus, to find the resultant of the four forces 
Fj^, F^, F^, F^, acting at the point 0, we draw the polygon 
0'F\F\F\F\ whose sides are parallel and equal to the four 
vectors representing the forces F^, F^, F^, F^. 




Fig. 13. 




By drawing 0'F\, we complete the polygon. This is the vec- 
cor resultant ; hence, the resultant of the four forces is R, parallel 
and equal to 0'F\. This vector polygon is usually called the 
force polygon. We can draw it in various ways, since the vectors 
may be combined in any order, giving the same resultant. The 



30 Elementary Mechanics 

resultant of a system of concurrent forces is then a unique force 
of definite direction and magnitude, acting at tlie common point. 

10. Equilibrium. — A particle is in equilibrium when all the 
forces which act upon it balance each other. It is evident that 
the resultant of these forces is zero, for if the resultant were of 
any magnitude the particle would move in the direction of this 
resultant. If the resultant is zero the force polygon closes. This 
condition enables us to calculate forces which are in equilibrium, 
for we can either construct the force polygon graphically and 
measure the unknown forces, or calculate their size by the geo- 
metrical relations of the figure. However, we can draw this force 
polygon onl}^ when we have not more than two unknown condi- 
tions; that is, two unknown magnitudes of forces, or two un- 
known directions, or a single force with both magnitude and 
direction unknown. It is well to note that at least two forces are 
required for equilibrium. A special case of the force polygon, 
which is of great importance, is the case of three forces in equi- 
librium. The force polygon is a triangle and the unknown parts 
of this triangle can be calculated by the geometrical or trigono- 
metrical relations of the figure. 

17. Illustrative Examples 

Example 1. — A particle is acted upon by two forces of 10 
pounds and 20 pounds respectively, making an angle of 60° with 




\o / 
— >• 



20 

Fig. 15. 

each other. Find the magnitude of the resultant and the angle 
which it makes with the 20 pound force. 



Statics of a Particle 



31 



Solution: Using the formulae of Article 8 we have 
i^2 3^ 102 + 20- + 2. 10. 20 cos 60°, 
i^- = 700. \\R = 26A5. 

10 sin 60° ^8.66 
25 



tan 6: 



.3464. 



20 + 10 cos 60" 
.•.6' = 19°6'. 
Note.— On Polyphase Slide Rule set 8.66 on C over 25 on D and read 
19° 6' (approximately) on T. 

Example 2. — Given a body of weight 10 pounds suspended 
from the ceiling by a string. The weight is pulled aside by a 
horizontal force F until the string makes an angle of 60° with 
the ceiling. Find F and the pull in the string. 



//////////////////// 




w=io 




>\c 



Fig. 16. 



Solution: The weight 10 pounds acts downward. The body 
is in equilibrium under the action of three forces : W downward, 
F horizontal, and P the tension in the string. Draw the force 
polygon, which in this case is the triangle ABC. W acts verti- 
cally with a length of 10 units; F acts horizontally but is of 
unknown length; P makes an angle of 60° with F. 
.•.i^=10cot60° = -Lo V3 = 5.77lbs. 

Note. — On Slide Rule set 30° on T scale, read result on C over 10 
on D scale. 

2 



.•.P=10csc60' 

Note.— On Slide Rule set 60^ 
10 on B scale. 



10 



V3 



11.54 lbs. 



on S and read result on A scale over 



32 



Elementary Mechanicis 



Example 3. — A weight of 60 pounds is suspended by two strings 
of lengths 12 and 9 feet respectively; the other ends of the 
strings are attached to two points in a horizontal line 15 feet 
apart; find the pull in the strings. 



//////////////////////////////////////////////////// 
A\ 15 90-6' XB 




60lbs 




Fig. 17. 



Fig. 18. 



Let Pi and P^ represent the pull in the 12- and 9-foot strings 
respectively. The strings and the line AB form a right triangle. 
From the force triangle (Fig. 18) 

P^ = 60 cos 6=60'^ P2 = 60sin^ = 60. i|, 
lo 15 



Pi = 36 lbs. 



P, = 481bs. 



Problems 

16. Find the magnitude and direction of the resultant of 
forces equal to 5 and 12 pounds respectively, acting at right 
angles. 

17. Forces equal respectively to 3, 4, 5, and 6 pounds act on 
a particle in directions respectively north, south, east and west. 
Find the direction and magnitude of their resultant. 

18. Two forces whose magnitudes are P and PV2 pounds act 
on a particle in directions inclined at an angle of 135° to each 
other. Find the magnitude and direction of the resultant. 

19. Find the resultant of two forces of 13 and 11 pounds 
respectively, acting at an angle whose tangent is M-. 



Statics of a Particle 33 

20. Two forces acting at an angle of 60° have a resultant equal 
to 2V3 pounds; if one of the forces is 2 pounds, find the other 
force. 

21. Find the resultant of two forces of 10 and 9 pounds respec- 
tively, acting at an angle whose tangent is f . 

22. Two equal forces act on a particle. Find the angle between 
them when the square of their resultant is equal to three times 
their product. 

23. An 8-pound force and an unknown force act at a point, 
and have a resultant of 12 pounds, which makes an angle of 30° 
with the 8-pound force. Find the unknown force to hundredths 
of a pound. 

24. The resultant of two forces of 8 and 10 pounds respectively 
is 5 pounds. Find, to the nearest minute, the angle between the 
8- and 10-pound forces. 

25. The resultant of two forces P and Q acting at right angles 
is R. If P be increased by 9 pounds and Q by 5 pounds, R 
becomes three times its former value, and makes the same angle 
with Q that it did before with P. Find P, Q, and R. 

26. A weight of 20 pounds is suspended freely from a fixed 
point by a perfectly flexible string. Find what horizontal pull 
applied to the body will move it out so that the string will make 
an angle of 30° with the vertical, and find the pull in the string. 

27. A weight of 4 pounds is suspended by a string, and is acted 
upon by a horizontal force. If in the position of equilibrium the 
tension of the string is 5 pounds, what is the horizontal force ? 

28. A weight of 10 tons is hanging by a chain 20 feet long. 
Find how much the tension in the chain is increased by the 
weight being pulled out by a horizontal force to a distance of 12 
feet from the vertical. 

29. The weight ir=100 pounds is supported at by two cords 
0^ and OB (Fig. 19). Using the 
" triangle of forces," determine the 
tension in OA, the angles between 
the three forces acting at being 
given. 

30. The sum of two forces is 18, 
and the resultant, whose direction 
is perpendicular to the lesser of the 
two forces, is 12; find magnitudes W=100 lbs 
of the forces. Fig. 19. 




34 



Elementary Mechanics 



31. Three concurrent forces are represented by lines drawn 
from the vertices of a triangle to the middle points of the oppo- 
site sides. Show that the resultant is zero. 

32. A cord whose length is 21 is tied to the points A and B 
in the same horizontal whose distance is 2a. A smooth ring 
upon the cord sustains a weight W. Find the tension in the cord. 

18. Types of Forces. — In the following articles we shall de- 
scribe briefly some of the common forces with which we are con- 
cerned in statics, and state the assumptions made regarding 
them. Forces thus considered are the weight of a body, the reac- 
tion of a surface, the reaction of pins on a member of a structure 
and the tension in cords and cables. 

The 'lu eight of a body is the force which pulls the body toward 
the earth. It acts in a vertical line. 

19. Action and Reaction. — In a chapter on the Dynamics of 
a Particle, which follows later, Newton's laws of motion are 
stated and discussed. However, it is necessary that we should 
at this time understand and employ his Third Law ; i. e., to every 




Fig. 20. 



action there is always an equal and opposite reaction: or, the 
mutual actions of any two bodies are always equal and oppositely 
directed. Thus, if one body pushes or pulls another it is itself 
pushed or pulled hy this other ivith an equal and opposite force. 
This law is a statement of fact, and has been veiified by compari- 
son of theory with observed results. 



Statics of a Particle 35 

Reaction of Surfaces. — If a body rests iipon a surface it presses 
against the surface because of its weight; and the surface reacts 
against the body. If the surface is smooth the reaction is normal 
to the surface at the point of contact. Thus^ if a body rests on 
a smooth inclined plane (Fig. 20), the reaction is R, the weight 
W acts vertically dowmvard. (Since the plane is smooth, seme 
other force, as F, is necessary to keep the body from sliding.) 

The student must have clearly in mind that this reaction is 
generally not equal to the weight. 

20. Law of Transmissibility. — If a force he applied to a rigid 
body at any point, the effect of this force will remain unchanged. 




Fig. 21. 

if its point of application he moved to any other position along 
the line of action of the force. Thus, in the figure AB is the line 
of action of the force F acting on the body M. No matter 
whether F is applied at A or i? or any other point along the line 
AB the effect of F will be the same. This is one of the most basic 
facts of mechanics. 

Reaction of Pins. — Many structures, such as bridges and roof 
trusses, are made of bars which are connected by means of pins. 
We shall assume for the present that these bars, or members as 



36 Ki.FMFN rAKY Mkouamos 

thov aro ofton onllod. aro vii^id bodios (^thai is, ilo not oliango their 
shape "i and that ihev are without weight. If the bar A is pnUod 
bv a foree applied at B and aeting aUnig- tlie axi;? of the bar, the 
pin (' reaets against tliis foree with pull equal in nnignitudo but 
opposite in dirooiiou. The bar in this ease is said to be in 
tension, and the foree aeting on the bar is ealled the tension in 
the bar. Likewise, if a push be exerted at />. there must be an 
equal and opposite pusli at (\ In this ease the bar is said to be 



Fiv;. 



in eonipression and the foree is ealled the eompression. This is 
an illustratiou of the law of transmissibilitv of foree. 

Tension in Cords or Cahhs. — A tlexible eord is one that nia\ 
be bent without ap}d_ving foree. .V eord in staiies will be eon.- 
sidered to be tlexible. iuextensible and weightless. For a eord to 
he in eqnilibriniu ovory point in it unist bo under the inthienee of 
two equal and opposite pulls or forees. Sinee a eord requires no 
foree to bend it, the direction of the tousiou along the eord is 
always parallel to the direction of the cord. Cords may carry 
tension in diltereiit directions by moans of pulleys or pegs. 

21. liiusiraiirv i^xamplcs 

E^wmpJe 7. — Let AOB (Fig. *^3) be a hinge fastened at the 
ends A and B, having a foree P acting vertically at the apex (>. 

It is required to tind the compresaiou in the arms -l() and BO. 
Drawing the force polygon (Fig. ^4^. we have the triangle O'A'B . 
which is isosceles. 

P 

llenee A - }> \^ • see f^ It is interesting to note that 

when the hinges are nearlv b.orir.ontal or t' is near i"'C the 



S'i'A'i'KJS Oh' A I^Airricii.io 



37 



fofcd ill l,li(! Iwo ;irms is cxI I'citicly l;ir<^('. Tiiis mcclijinlsrn is 
ciillcd a io^^lo joiiiL ;iii<l is used to obtain gr(!jit prossuro; an 



.A 





Fkj. 24. 

illiisti'iitioii of its iis(! is round in iiicclianisFUs used in llu; ci'ush- 
iii^- of rock. 

/'J.niwjilr, :'L \H no be n. bar of Icn-tb I tcct and A li and AG 
be (d' f) feet and (> feet, iH'spcct i vcl y, 
with a load of a. hundred pounds 
aciino- at Iho point (). Tho bar IK) 
is kept from turning- by a cable) A(). 
'V\\(' pi-()bl('in is lo (ind th(! tension 
in llie eabK; and (he; e()ni[)i'ession in 
th(« bar. 

Shouhl \\v. now di'aw a, force poly- 
gon (which in Ibis case woidd be a 
trian<4'le) it would necessarily !)(> 
similar to the given triangle AliO, 
since its si(l(\s would be resp(>ctiv(dy 
parallel to the sides of AliO. We 
(;an use any unit ot IcMigth whalever 
to r(>present a. unit of forc(>; henc(>, 
we can use the lengih 5 oi- AJ> (o 
represent the force of 100 pounds 
aetiiig vertically. In oIIkm- words, (h(> triangle A HO can be used 
as the force triangle; thei-eroi-e we have 
•00^r.„>^6V,, 
5 (; 4: ■ ■ 

Hence, T.io^l'^O pounds, and (J,(q — ^0 i)oun(ls. 

No'PK. -Sot 100 on C ovov 5 on I), on C read T,,, over G and C^(, 
over 1. 




Fin. 25. 



38 



Elementary Mechanics 



Example 3. — Suppose that a block oi' weight IV rests on a 
smooth iucliiiod ph\iK% aiul we wish to (iiul tlie force P necessary 




Fk;. '26. 

to keep it from sliding-, together with the normal reaction A^. 
Constructing the force polygon ODW. which is a right triangle, 
we have 

DO = r=]Y-s\uO and DlT = iA^= 11" -cos ^. 

Example J/. — A string BAC passes over two smooth pegs B and 
C and carries weights IFi^-rl: and T7o = o at the ends, as shown 
in the figure. What weight, P, suspended from a point A, on the 
string, will make ABAC = 120'' ? Draw the force triangle TIDE, 
and we have 

P- = 4:= + 3--2 • -f • 3 cos t)0°. 
/.P=3.Gllbs. 

H . 




Wr3 



'W,=4 



Fig. 27. 




Statics of a Particle 



39 



Example 5. — Let BO be a bar 
with a load of 100 pounds acting at 
the point 0. The bar is kept from 
turning by a cable AO. To find the 
tension in AO, and the compression 
in BO. It will be seen that the tri- 
angle AOB is the force triangle, with 
the side AB corresponding to the 
100-pound weight. We have then 

100 ^ r,o ^ Co 

sin 15^ sin 30° sin 135°* 
Hence, 

2^40 = 193 Lbs. 
and 

(750 = 273 lbs. 




Fig. 29. 



Note.— Set 100 on A over 15° on S and on A read 193 over 30° and 
273 over 45°. 

Problems 
33. A weightless rod AC hinged at one end A so as to be free 



B 






6 


\ 






8 


\- 


A 







w 




Fig. 30. 



Fig. 31. 



Fig. 32. 



to turn in a vertical plane, is held in a horizontal position by 
means of the chain BC (Fig. 30). If a weight ^Y be suspended 



40 



Elementary Mechanics 



at C, find the thrust P in AC and the tension T of the chain. 
Assume AC equals 8 feet and AB equals 6 feet. 

34. Find the compression in the bar AB (Fig. 31) and the 
tension in the cord BC. 




Fig. 33. 



35. Find the tension in the cord AB and the compression in 
the bar BC, due to a load of 2 tons at B (Fig. 32). 

36. If a toggle joint ABC has the end A hinged, find the hori- 
zontal force F in terms of the load P at the apex (Fig. 33). 




300lbs 

Fig. 34. 



lOOIbs 
Fig. 35. 



37. The 300-pound weight of Fig. 34 is attached at C to the 
cords BC and AC. Find the tension in BC. 

38. The 100-pound weight in the figure is attached at C to 
the cords BC and AC. If the tension in BC is 70 pounds find 
the angle 6 to the nearest tenth of a degree (Fig. 35). 

39. A cord is attached to a fixed point and passes over a smooth 
pulley on the same level with the point and 10 feet distant (Fig. 



Statics of a Particle 



41 



36). A smooth ring of weight i(;. slides on the cord; the cord is 
kept tight by a weight W on the free end. Find a deptli to 

W 



which the ring will sink when (1) iv = W; (2) when iv = 



3' 



40. Three pegs form an equilateral triangle ; if an india-rubber 
band be stretched around the pegs, and if the pull in the band be 
equal to a weight of 10 pounds, find the resultant pressure on each 
peg. 

41. A wheel weighing W pounds is about to roll over an 
obstruction. The radius of wheel is R and the height of the 
obstruction is h. Find the horizontal force through the center 
necessary to start the wheel (Fig. 37). 



lOft 




Fig. 36. 



42. Two rafters making an angle of 120° support 112 pounds 
at the peak. Find the compressive force on each rafter. 

43. Show that if a picture is hung from a nail by a cord 
secured to two rings in the top of the frame, the shorter the cord 
the stronger it ought to be. Would it be possible for the cord to 
remain straight when placed over the nail? 

22. Resolution of Forces. — We have seen that a number of 
forces may be replaced by a single force. The converse of this 
is also true ; we can replace a force by a number of forces. While 
the former case is unique in solution, there being only one resul- 
tant, the latter is not unique as there are an infinite number of 
ways a force can be resolved into component forces. The most 
important case of resolution of forces is that which gives us com- 



42 



Elementary Mechanics 



ponents along two mutually perpendicular lines. Let AB he a. 
force, then by the parallelogram law it is the resultant of the two 
perpendicular forces AC and AD. 

Hence 

AO = AB cose, 

AD = BC = AB sin 6 = AB cos </,. 

We have then the rule : The component of a force along one 
of two perpendicular lines is the product of the force and the 




cosine of the angle between the line of action of the force and the 
given line. 

23. Algebraic Method for Finding the Resultant of a System 
of Concurrent, Coplanar Forces. — If a number of coplanar forces 
act upon a particle then they can all be resolved along any pair 
of mutually perpendicular lines, and the algebraic sums of the 
components will be the components of the resultant. If a system 
of coordinate axes x and y are taken, and X is the sum of the com- 
ponents along the :c-axis and Y along the ^-axis, 



and tan (/)=_• 
2l 



R-coscf) = X, i? =X sec <^ = Fcsc </>, 

where </> is the angle which the resultant R makes with the a:-axis. 



Statics of a Particle 



43 



In calculating tlie resultant due regard must be taken of the 
signs of the trigonometric functions. It might also be noted that 
a force equal and oj^posite to R would put the system in equi- 
librium; this balancing force is called the equilibrant. 

Example. — Let us consider seven coplanar forces acting on a 
particle, as follows: f V3, 4V2, 2, 3, 6, 5^, and 8 pounds, mak- 
ing with a fixed line in the plane these angles: 0°, 45°, 120°, 
150°, 240°, 270°, and 300°, respectively. Eequired the resul- 
tant force and the angle it makes with the a;-axis. 

fV3"cos 0° = |V3 
4V2cos 45° =4 

2 cos 120°= -1 _ 

3cosl50° = -|V3 

6 cos 240° r= -3 
5^ cos 270° = 

8 cos 300° =4 



X = 4 

IVjsin 0° = 
4V2sin 45° =4 _ 
2sinl20° = V3 
3 sin 150° =1 _ 
6sin240°=-3V3 
5i sin 270° =-5^ _ 
8sin300°=-4V3 




r=-6V3 



VZ^H-r-=Vl6 + 108 = 2V31=:11.14, 

Y o - 

arctan ^ =arctan— |V3, 



<^ = 291°3', 
Note.— To find 8 cos 300° set 30° on ^S and read result on B under 8 
on A; to find 8 sin 300°, set 60° on >S' and read result on B under 
8 on A, etc. 

Set 1 on C scale under 3 on A scale, move the runner to 3 on C 
scale; then set 2 on C scale under runner, read 21° 3' on T scale. 
.•.0=270° + 21° 3' = 291° 3'. 
Set 21° 3' on /S and read 11.2 on A scale over 4 on B scale. 



44 



Elementary Mechanics 



24. Algebraic Conditions for Equilibrium. — If a system of 
forces acting -apon a particle is in equilibrium, the resultant is 
zero. If the resultant is zero its resolved part along any line must 
be zero. We have then, for a system of coplanar forces in equi- 
librium, the following important law : The algebraic sum of the 
components along any line must equal zero. 

Since we can resolve a force along any line, it is well to select 
the line (or lines) in such a way as will tend to simplify the 
algebraic work. This may often be done by resolving in a direc- 
tion perpendicular to one of the unknown forces. 



Example 1. 
body W, weighin 



Given the forces shown in Fig. 40, acting on a 



hundred pounds, and resting on the smooth 



inclined plane AB. Find the force, P/ just sufficient to keep 



50lbs 
45 




Fig. 40. 



the body from sliding; also the normal pressure or reaction N. 
Let us resolve along the plane and perpendicular to it. 

We have then 



Also 



Z = = P . cos 30° - 50 . cos 45° - 100 • cos 60° ; 

.•.P = 98.53 1bs. 

r = 0z=iiV + 50.sin45°+P. sin 30°-100. sin 60' 
/. iV = 86.60-49.27-35.36 = 1.97 lbs. 



Statics of a Particle 



45 



In solving for N, we might have resolved perpendicular to P, 
and thus secured an equation in N alone; but this would have 
proved less simple than the method used. 

Often we have problems in which it is necessary to solve suc- 
cessively for the various unknowns. If a series of particles or 
joints are connected by a system of ropes or bars^ we may start 
our solution at one of the joints. The joint must be such that 
not more than two of the forces acting on it are unknown. This 
joint is in equilibrium^ and hence we can solve for the unknown 



A \ 





Fig. 41. 



2 tons 

Fig. 42. 



forces acting upon it. This will give us known forces and will 
usually enable us to proceed to a neighboring point and solve its 
system. It must be kept constantly in mind that every joint is 
in equilibrium under the forces which act upon it and that the 
force exerted at one end of a member is equal and opposite to that 
exerted at its other end. 

Example 2. — Let Fig. 41 represent a crane, supporting a load, 
at the point E, of two tons. To find the stresses in the members, 
start at the point E. The forces acting at this point are the 
weight, and the stresses in the two members BE and ED. The 
stress in ED must evidently act upwards along ED, to balance 
the downward pull of the load, since the stress in BE has no 



46 Elementary Mechanics 

vertical component. Similarly, the stress in BE must act upon ' 
E towards the right. Eesolving horizontally and vertically, we 
have 

X = = BE-ED^cos60% 

Y = 0=:ED.sm60°-2; 

4 
.'.ED—^r=^ (tension), 
V3 

2 
BE=^ —=. (compression). 

Note that in the case of tension the stress pulls on the pin or 
joint at E ; whereas in the case of compression the stress pushes 
against the point E. 

Let us now proceed to the point J). Since the member BE is 
in equilibrium the force acting at J) must be equal and opposite 

A.. ° 




Fig. 43. 

in direction to the force acting at E. Thus, the forces acting at 
I) are shown in Fig. 43. Eesolving horizontally and vertically, 
as above, we get 

Z = 0=: -^D + 5D.cos60°+ Acos60°, 

V3 

AD-^^^, (1) 

also ^ V3 ' 

rzr:0=:5i).sin60°- isin60% 

4 V3 

. . BE— — = (compression) • 
and from (1), 

AD— — =^ (tension). 



Statics op a Paeticle 



47 



Prohlems 

4:4:. Four forces of 8, 26, 20, and 16 pounds act on a particle; 
the angles between the S-ponnd force and the other forces are 
respectively 20°, 65°, and 70°. Find graphically the resultant. 
(Let 1 inch represent a force of a 4-pound weight.) 

45. Forces of 11, 16, 13, and 10 pounds act at a point in 
the plane XOY, and make angles of 30°, 90°, 180°, and 310°, 
respectively, with OX. Find the resultant of these four forces 
to hundredths of a pound and the angle (to the nearest tenth of 
a degree) which the resultant makes with OX. 

46. A captive balloon capable of raising a weight of 400 pounds, 
is anchored at a height of 400 feet by a rope 500 feet long. Find 
the pull in the rope and the horizontal pressure of the wind on 
the balloon. 



W=60Ibs 




Fig. 44. 



47. The tow-line of a canal boat is 150 feet long from the 
fastening on the boat to the harness of the mule. The tow-line 
includes a spring-balance which reads 90 pounds. The mule 
walks 3 feet from the edge of the water, and the attachment of the 
tow-line to the boat is 12 feet from the bank. Neglect the sag of 
the line. 

(a) AYhat is the force tending to turn the bow toward the bank ? 

(b) What is the effective pull on the boat in the direction of 
the desired motion ? 

(c) What is the percentage of the former to the total pull? 

(d) How would the sag of the line modify these results? 

48. W is held at rest on the smooth plane AB by the horizontal 
20-pound force, shown in Fig. 44, and by the force P, and the 



48 



Elementary Mechanics 



reaction of the plane on W. P acts parallel to AB. Find the 
force P, and the reaction of the plane AB on W. 

49. A 160-ponnd weight rests on the smooth plane AB, and is 
held in equilibrium by two forces of 80 pounds each, acting paral- 




FiG. 45. 



Fig. 46. 



lei to AB and AC, respectively. Find the inclination of AB to the 
horizontal (to the nearest min.) and the reaction of AB on W 
(Fig. 45). 



4? 24 Ions 




Fig. 47. 



50. A weight W rests on a smooth plane and is kept in position 
by a string passing over a pulley and attached to a weight of P 
pounds hanging freely (Fig. 46). Find all the forces when 



(TATICS OF A PaKTICLE 



49 



(a) 
(b) 



30°, ^ = 30°, P = 15 lbs. 
40% ^ = 15°, 17=10 lbs. 



51. The 50-foot spar AB, used as a derrick, is supported by two 
back stays, BH and BK, each 100 feet long. Given DA — 4:5 feet, 
AC = 30 feet (Fig. 47), find the tension in each back stay when a 
weight of 24 tons hangs at B (neglect weight of spar). 

25. Friction. — When we attempt to slide one body over another 
there is resistance offered to the motion because of the rouo-hness 
of the two bodies. This resistance is called friction. There are 
three fundamental laws of friction with which we shall be con- 
cerned : 

(1) The friction between two bodies is directly proportional to 
the pressure. 

(2) The amount of friction for any given pressure is indepen- 
dent of the area of contact. 




lOOIbs 

Fig. 48. 



(3) The direction of the friction is tangent to the surface upon 
which a body tends to move and is opposite to the direction in 
which the body tends to move. 

These laws do not apply when the pressure is so great as to cut 
one of the sliding bodies. The ratio of the friction to the pressure, 
that is, to the normal pressure, is called the coefficient of friction, 

W 

and is designated by the letter /x. Thus /x= — . 

The friction F is the total force that the roughness of the 
bodies can offer to sliding. It must be remembered that the fric- 
4 



50 



Elementary Mechanics 



tion or resistance that a body offers to movement is eqnal to the 
force appHed, in the direction of motion, nntil the friction reaches 
its maximum vahie. Thus, if a block weighs 100 pounds and the 
coeiticient of friction is -|, the total force tliat the friction can 
exert is 50 pounds. 

If the force P is greater than 50 pounds the block will slide. 
On the other hand, if the force P is only 2 pounds the friction is 
only 2 pounds. Ordiiuirily when we use the word friction we 
mean the maximum or so-called '' limiting friction." 

26. The Angle of Repose. — If a body of weight W is placed 
on a rough inclined plane, as in Fig. 49, the friction tends to 
keep the body from sliding. If the angle 9 is increased until 
slipping just begins we have 



I 



tan(/>= -^, 



W sin 



tan<9: 



and 61: 



]V cos 

or the tangent of this angle is equal to the coefficient of friction. 
This angle, which the inclined plane makes with the horizontal 




Fic. 49. 



at the instant the body begins to slide, is called the angle of repose. 
The coefficient of friction is usually determined by placing a 
body on an inclined plane and finding the angle of repose. 



Statics of a Particle 



51 



"When a body moves (or is just on the point of moving) on a 
ronofh surface the reaction is no lono-er normal to the surface 
but acts at an angle equal to the angle of repose from the normal; 
it always acts opposite to the direction of motion. 



27. 



Illustrative Problems on Friction 



Example 1. — A block is being acted upon by a force P which 
is about to slide the block along the floor. 



50lbs 




The weight of the block is 100 pounds and the coefficient of 
friction is \ There is also another force of 50 pounds pulling 
back on the block. To find the value of P that will just cause 
the block to move towards B. Let the reaction of the plane on 

the block be N, and the friction be P. Then ^= t- Resolve 

horizontally and vertically 

Z = = P.cosl5°-o0.cos30°-^, 

r = 0=P.sinl5°+xA^ + 50.sin30°-100; 
.*. N=2hM lbs., and P = 70.2 lbs. 

Example 2. — An unknown weight 17 and a weight of 20 
pounds rest on a double inclined plane as in the figure ; the angles 
which the two planes make with the horizontal are 45° and 30°, 
respectively. The weights are connected by a string passing over 
a smooth pulley. The coefficient of friction between the 20- 



52 



Elementary Mechanics 



pound weight and the plane on which it lies is J VF; the other 
plane is smooth. Find the weight W and the tension in the 





Fig. 51. 

string, (a) when the 20-pound weight is about to move up the 
plane; (b) when it is about to move down the plane. 

At the point A there are three forces acting; W, vertically; 
T, the tension in the string, which acts 
along the plane; and N, the normal 
reaction of the plane. This is illus- 
trated in Fig. 52. At B, however, we 
have four forces acting: 20 pounds ver- 
tically, T along the plane, TV' normal 
to the plane, and F the friction along 
the plane. F acts opposite to the ten- 
dency of motion; i.e., (a) if the 20- 
pound weight is about to move up the 
plane F acts down the plane (Fig. 53) ; 
and (b) if the 20-pound weight is about 
to slide down, F acts up the plane (Fig. 
54). 
In Fig. 52, resolving perpendicular to N, we get 

T=:W.sin45°= — 
V2 
In Fig. 53, resolving perpendicular to N', 
T = FH-20.cos60°=iP+10; 
and resolving along N', 

iV' = 20.cos 30° = 10V3. 
But 

F = fjiN'=^^ •10V3 = 5lbs. 



W 

Fig. 52. 



(1) 



. T=10 



and from (1), 



2V3 
5 = 15 lbs. 



14^ = rV2 = 15V21bs. 



Statics of a Particle 



53 



These are the values of 17 and T when the 20-poiind weight is 
about to move up the plane. In the same way, usinor Fig. 54, and 




N 



f'V 



60 



20lbs 
Fig. 53. 







20lb5 
Fig. 54. 



resolving perpendicular to N' , we have r + i^ = 20 • cos 60° =10; 
.•.T=10-i^. _ 

Resolving along N' . . . . iV' = 10V3. 
Therefore 

i^=5lbs. 
T = 5 1bs. 
and from (1), as before, _ _ 

Ty = rV^=5V2lbs. 
These are the values which obtain when the 20-pound weight 
is about to move down the plane. 

Problems 

52. A block, weight 100 pounds, resting on a horizontal plane, 
is pulled by a force P which makes an angle of 10° with the 
horizontal plane. If the coefficient of friction is 0.2, how great 
must P be to start the block ? 

53. If a block of weight W rests on a horizontal plane, find the 
angle that will require a minimum force, P, to just start the 
block in motion. The coefficient of friction is /x. 

54. If in problem 52 the force P is pushing against the block, 
find P. 

55. Solve problem 52, using 25° and 0.3 in lieu of 10° and 0.2, 
respectively. 



54 



Elementary Mechanics 




Fig. 55. 



5Q. Show that the force P (Fig. 55), inclined at an angle </> 

to the plane, that will (a) just move 
the weight up the plane, (b) just 
prevent it from sliding down the 
plane, is 

, . p_ T7sin(a + ^) 

^^^ ^~ cos(c/>-^) ' 

(b) P= TFsin ia-O) 

cos (cA + ^) ' 

where is the angle of friction 

(repose). 

57. Show that the least values of P in the preceding problem 
occur when cf) = in (a) and when </>= — ^ in {h) ; i. e., when 

(a) P = Tf sin (a + ^), (b) P=]y sin (a-^). 

58. In Fig. 56 the weight W is raised by a horizontal force P. 
If the only friction is between the surfaces of the wedge and the 
weight, prove that the value of P just sufficient to raise the weight 
is 

P=Ftan (a + ^), 

where is the angle of friction. 





w 




'/ 
/ 


- 


- 




^ 


^^-^^^ 


\ 
I 


^ p 


////////^///v//////////// 


//}///////// 


y/y/y/y////////// 


'////////J/^/ 



Fig. 56. 



59. A body, whose weight is 20 pounds is about to slide along 
a rough horizontal plane, under the action of a force whose 
inclination to the horizontal is 45°. Find the magnitude of the 
force, \x being 0.3. 

60. A weight W rests in equilibrium on a rough inclined plane, 
being just on the point of slipping down. On applying a force IF, 



Statics op a Particle 55^ 

parallel to the plane, the weight is just on the point of moving iip. 
Find the angle of the plane and the coefficient of friction. 

61. A weight of 30 pounds is resting on a rough horizontal 
plane and can just be moved by a force of 10 pounds acting hori- 
zontally; find the coefficient of friction and the direction and 
magnitude of the resultant reaction of the plane. 

62. Two rough bodies ]T\ and Wo rest upon an inclined plane 
and are connected by a string parallel to the plane. If the coeffi- 
cient of friction is not the same for both, determine the greatest 
inclination consistent with equlibriuni, and the tension of the 
string. 



'////////////////////////// 



F 




Fig. 57. 



63. A force F acts upon a sliding member A (Fig. 57), which 
is connected to a bar AF , and this bar is connected to a fly wheel. 
Find the force Q, perpendicular to the radius, that will keep the 
wheel in equilibrium, if the coefficient of friction at A is ^a. 

Review Prohlems 

64. A force equal to 10 pounds is inclined at an angle of 30° 
to the horizontal ; find its resolved parts in a horizontal and ver- 
tical direction respectively. 

65. Find the resolved part of a force P, in a direction making 
(1) an angle of 45°, (2) an angle equal to the cos"^ || with its 
direction. 

66. A truck is at rest on a railway line, and is pulled by a 
horizontal force of 100 pounds, in a direction making an angle 
of 60° with the direction of the rails. What is the force tending 
to urge the truck forward ? 



56 



Elemextaey Mechaxics 



6T. Find the components of a force of *50 ponnds along two 
directions, making angles of 60^ and 45^ with it on opposite sides. 

68. If a force P be resolved into two forces making angles 
of 45^ and 15^ with its direction, show that the latter force is 

PVl 

69. If a force F = '20 ponnds be resolved into two miitnally 
perpendicular components, one of which is donble the other, lind 
the greater component, and the angle it makes with F. 

70. A force of 2b poitnds acting vertically npwards is the 
resultant of two forces, one being horizontal and eqnal to 10 
poimds. What are the magnitude and direction of the other force ? 

71. Find a horizontal force and a force inclined at an angle 
of 60^" with the vertical whose resultant shall be a given vertical 
force F. 

7C. (^a) Three forces acting at a point are in eqnilibrinm. If 
they make angles of 120^^ with one another, show that they are 
eqnal. 

(b) If the angles are 60"^, 1lHV\ and 150"^, in what proportions 
are the forces? 

73. Three forces acting on a particle are in eqnilibrinm; the 
angle between the first and second is OO'', and that between the 
second and third is 120^ ; find the ratio of the forces. 

74. Forces eqnal to 5 IF. 12 IF. and 13 IF acting on a particle 
are in eqnilibrinm ; find the angles between their directions. 

75. Constrnct geometrically the directions of two forces 21F 
and o^y which make eqnilibrinm with a force of 4Tr whose direc- 
tion is given. 

76. The sides AB and AC of a triangle ABC are bisected 
in D and B. Show that the resnltant of forces, represented by 
BE and DC is represented in magnitude and direction by f BC. 

7 7. Two forces act at an angle of 120"". The greater is rep- 
resented by SO. and the resnltant is at right angles to the lesser. 
Find the latter. 

78. If one of two forces be dotible the other, and the resnltant 
be eqnal to the greater force, find the angle between the forces. 

79. Two forces acting on a particle are at right angles and 
are balanced by a third force, making an angle of 150^ with one of 
them. The greater of the two forces being 3 pounds, what must 
be the valnes of the other two ? 



Statics of a Particle 



SO. AB i< a 13-foot boom, J. is a small hinge, and BC is a 20- 
fooT giiT i^Fig. 5S). Find the thrust in AB and the tension in 
BC, when 33 tons hano^s from B. 




33 tons 



51. A weight of 42 ponnds is supported hy two strings of 
lengths 2b and IT inches, which are tied to two pegs in the same 
horizontal line at a distance of 2S 

inches apart; find the tension in 
the strings, and the greatest hori- 
zontal force which can be applied 
to the weight without disturbing 
the equilibrium . 

52. A weight of 2 TO poimds 
is suspended from a crane of 
dimensions shown in Fig. 59. It 
is required to find the tension 
in AB and the compression in 
BC. 

53. A small ring is held at the 
center of a hexagon by six cords, 
all in the same plane, and attached 

each to a separate yertex of the hexagon. The tensions of four 
consecutiTe cords are 2, T, 9, and 6 pounds, respectively. Find 
the tension in the remaining two cords. 

54. P, Q. and B are weights attiiched to strings, two pass over 
pulleys and the three strings are knotted together at (^see 
Fig. 60). Find the tension in the strings when 

(a^ Q = 10 lbs., a = lbO\ 3= 60'; 




270 lbs 



Fig. 59. 



(b) P= 5 lbs., a = 120"^ 

(c) B= 6 lbs., a = 100" 



,8= 90'; 
/8 = 140°. 



58 Elementary Mechanics 

85. P, Q, and R are weights attached to strings knotted at 0, 




Fig. 60. 



and passing over two pulleys (see Fig. 60). Find the tensions 
in the strings when 

(a) P:= 15 lbs., a = 90°, ^ = 120°; 

(b) g = 2000 lbs., a=70°, /? = 150°; 

(c) R^ 2.5 lbs., a = 60°, /? = 150°. 

86. On the opposite sides of a straight line AB two equilateral 
triangles with vertices C and D are described, and a force P acts 
along AB. Resolve this force along the directions AC and AD. 




lOibs 

Fig. 61. 



87. AD is the perpendicular from the angle A of an equilateral 
triangle ABC on the opposite side BC. If the resolved parts 



Statics of a Particle 59 

along AC of the two forces P and Q which act along AB and AD 
respectively are equal to one another, find the ratio of P to Q. 

88. Three strings are knotted together into the shape of an 
isosceles right-angled triangle (Fig. 61). A weight of 10 pounds 
is suspended from the right angle, and the hypotenuse is kept 
horizontal by two strings attached to its ends, which are both in- 
clined at 30° to the horizon. Find the tension of each string. 

89. Let B be the effective piston pressure of a steam engine 
and (f> the angle between the direction of motion of the piston 
and the connecting rod at any moment. Show that the thrust 
in the connecting rod is R sec 4> and the pressure on the guide 
bars is R tan </>. For what position of the crank is the pressure 
on the guides greatest ? What is the greatest pressure when the 
connecting rod is four times as long as the crank? 

90. The force R sec (^ transmitted by the connecting rod is 
resolved at the crank pin tangentially and radially. Determine 
the tangential component which is the effective turning force 
and show that, if R be represented by the length of the crank, 
this tangential component is represented by the intercept made 
by the connecting rod on the radius of the crank circle which is 
perpendicular to the motion of the' piston. Show also that when 
the connecting rod is long the tangential force is approximately 

= R (sin $-\- ^— sin 20), where is the crank angle and m the 

ratio of the length of the connecting rod to that of the crank. 

91. An anchor weighing 4000 pounds is supported by two 
tackles from the fore and main yards of a vessel, making angles 
of 30° and 45° respectively with the vertical. Find the tension 
in each tackle. 

92. A barrel 4 feet long, weighing 500 pounds, is hoisted from 
a ship's hold by means of a pair of can-hooks 52 inches long. 
Find the tension on each leg of the can-hook. 

93. Show that the least force which will move a weight W 
along a rough horizontal plane is W sin (/>, where <f) is the angle 
of friction. 

94. ABCD is a parallelogram ; a particle P is attracted toward 
A and C by forces which are proportional to PA and PC, respec- 
tively, and repelled from B and D by forces proportional to PB 
and PD; show that P is in equilibrium wherever situated. 



60 



Elementary Mechanics 



95. Two forces make an angle of 60° with each other, and 
their resultant is 14 pounds. If one force is 4 pounds greater 
than the other, find the two forces and the angle (to the nearest 
minute) which their resultant makes with the greater force. 

96. Find the thrust on the boom CB of the Fig. 62, due to the 
load TF = 3000 pounds. 




3000lbs 



Fig. 62. 



97. A weight of 30 pounds is resting on a rough horizontal 
plane and can be just moved by a force of 10 pounds acting i 
horizontally; find the coefficient of friction and the direction i 
and magnitude of the resultant reaction of the plane. 



Forces Acting on a Kigid Body 61 



CHAPTER III 
FORCES ACTING ON A RIGID BODY 

28. Definition.— A rigid body is one which experiences no 
change of shape when acted upon by forces however large. Or, 
mathematically speaking, a rigid body is one any two points of 
which remain the same distance apart, no matter how great the 
forces may be that act upon it. No such bodies exist in nature, 
but some may be regarded as rigid so long as the forces acting 
are not too great. 

Thus if moderate forces are applied to a cube of wood or steel, 
no appreciable change occurs in its size or shape, whereas, if the 
same forces are applied to a piece of rubber or putty, deformation 
takes place. Within suitable limits, then, we can regard the 
steel and wooden cubes as rigid bodies. 

This is an illustration of a common practice in applied mathe- 
matics, namely, that of making, for the sake of simplicity, an 
assumption which is known to be not strictly true, but which 
is near enough to conformity with the facts of nature to give us 
results which are very nearly correct. In many cases these results 
serve well enough for all practical purposes. In other cases, 
where they are not sufficiently accurate, they furnish a first 
approximation, that is, a starting point for more refined investi- 
gation. Thus it is convenient and practical for many purposes 
to regard pieces of wood, steel, etc., as rigid bodies. 

29. A rigid body can move about in space without changing the 
direction of any line in it. This is a motion of translation. It 
can turn about a fixed axis. This is a motion of rotation. It can 
be shown that any displacement of a rigid body is one or the other 
of these two or a combination of them. The body has no motion 
when any three points, not in the same straight line, are fixed. 



62 Elementary Mechanics 

For, let .:l. B, C, bo throe points not in the same straight line. 
If we fix A and 7^'. the body oan have only one motion, that of 
rotation abont tJie line AB, sinee by hypothesis the body is rigid 
and the distanee between AB and any other point F remains 
unchanged. This point P must describe a circle with axis AB. 
Thns, if C is not in line with AB, it mnst describe a circle about 
the Hue AB. Bnt if C is fixed this does not happen. Hence no 
motion can take {^laoo and the body is fixed. 

30. If a system of forces acts upon a rigid body, we frequently 
find that it may be replaced by a simple system or a single force 
without any change in its eiTect on the body, that is, in its ten- 
dency to produce translation or rotation. Such a single force or 
simplified system which is equivaleut to the given system is 
called the residtant of the system. This use of the word '* resul- 
tant '^ is a generalization of its use as applied to a particle in 
the preceding chapter. There, of course, the only possible effect 
was a teudoucy to produce translation, and the only law of sub- 
stitution was found to be the parallelogram law. For a rigid 
body we find it sutficient to make, in addition to this, the two 
following assumptions : 

(1) That the effect of a force on a rigid body is unchanged if 
its point of application is moved to any other point of the body 
in the lino of action of the force. 

(•?) That the effect on a rigid body is unchanged if two equal 
and opposite forces, acting at any point of the body, are removed 
or introdueod, 

31. The Moment of a Force About a Line. — Definition: (1) 
If a line L is perpendicular to the line of action of a force F, 
the mouuuit of the force with respect to the line L is the product 
of the force and the length of the common perpendicular between 
the two lines. Thus, in the Fig. i)3, the moment of F about an 
axis tlirough () and perpendicular to the plane of the paper is 
FxAO. 



Forces Acting on a Eigid Body 



63 



(3) If the lino L is not perpendicnlar to the line of action 
of the force F (Fig. i)!), resolve the force into two components, 
one parallel and the other perpendicular to the line L. Then 
the moment of the force is the moment of its perpendicular com- 
ponent h\ about the line L. 

L 






VJ 


X- 


/ 







Fig. 64. 

In both cases tlie moment measures the tendency to produce 
rotation of a body about the line L. In the second ease the paral- 
lel component has no tendency to produce rotation about the line 
L. The line L is called the axis of moments. The unit of 
measurement for the moment of a force must involve the units of 
force and distance, and accordingly the unit of moment of force is 
the pound-foot, the ton-inch, or the like. Thus a force of 10 
pounds acting perpendicular to a certain line and at a distance 
of 5 feet from that line has a moment of 50 pound-feet about 
that line as a_\is. 

32. When we consider moments of forces in one plane only, 
we shall speak of the moment of 
a force about a point. The 
moment of a force about a point 
is equal to the product of 
the magnitude oi the force and 

the perpendicular distance from O 

the point to the line of action of 
the force. This perpendicular 



Fig. 65. 



64 



Elemkntaky Mechanics 



clistaiU'o is known as tlio arm o( tho (owe. Tlu^ point (^ is caHod ■ 
tho (.ontor o( nuMnonls. Tho piano ilotorniinoil h\ iho point (^ 
anJ tho line o( action of tlio I'oroo is tlio [>lano ot' nuunonts. Tlio 
axis of nuunonts is a lino tliron^li (K {HM-ptMulioular to tho piano 
o{' tncnnonts. Thon, aooorilini'; io tho ilollniiion, tho nioniont o( F 
ahout oqnals /•' • a ( 1m_;;- tio) . 

33. Sig'ii of a Moment. — If tho totulonoy o{ tho tnoniont is to 

turn tho hodv oonntor olook\\iso, tho nioniiMit is positivo; if olook- 
wiso, it is nogativo. 

34. 'FlicorcDi.- Tho moniont about a point of a ioxco F 
artiiu;-at a point .1 is Oipial to (M nuilti[>lu\l hv tlio ooin[ioiiont 
o:f F m a diiwtion por[ioiulioiilar to (^.l. 




Fii\. 60. 



Let PA be the lino o( aotion o( a foroo F in ibo plane of the 
paper actino- on a panioK^ at .1 (Im^:'. iUO . Praw a porpendieular 
ON from anv poiiu () to /M. Lot t) bo tho angle AON. 
Draw - 1 ."^^ pcM-poiulioiilar to (^.L 

Thon the luoiuont of I'oroo F about — F'ON 

-F 'OA COS e 
=zOA'F cos $ 
==0^4 • oomponont o'i F 
alono- *>M. 



li'OKOI'lS ACTTNC ON A RkUD HoDY 



65 



35. 'r/huircm. 'V\]o sum of ilu> monuMits about any poiiil 
o\' ;iiiy iiniiilxM' o\' i'()iiciii-i-i'iit I'orci's is (>((iinl to the monuMiL al)i)iiL 
( ) o\' I liri r i'(\siill;m( . 

I.vl A',. /''.,. A',. .... h(> Ihc forces acting at A, and R tluM'r ro- 
siiliaiii. lu'l'orriiii;- to tlu> above figure (Fig. (U)), we liave to 
sbow that ()A • eoinponeiit of 7'\ along AS-\-OA. • coniponeiit of 
/''.. along /1*S'-|-. . . . = 0/1 • component of R along AiS. 

Dividing through by OA we have only to show that the co^n- 
poncnt of A* along /1*.S' is equal to th(» sum of the com])onents of 
/'',. A'... A'.. .... ah)ng AS. which is l<iu)\vn to bo true. (See 

Art. '/;>.) ' 

If a [tarticK^ is in (Hpiilibrium under the action of any nnnd)er 
o{' foi('(>s, the iH\^ultant o\' all tliese forces must be zero. The sum 
of {\\v moments taken about any point whatever is equal to the 
monuMd o\' the n^sultant and is therefore zero, 

ll(Mic(\ wluMi a particle^ is in equilibrium nndiM* the action of 
any number of c(UU'un\Mit forces, the sum oi' the monuuds of 
ilw^c {'ovcci< about any point whatever must vanish, 

36. Non-Parallel Forces.— L(>t r and Q (Kig; OT) he two 
non [)arallel t'oplanar forces acting ou a rigid body at .L and 7). 




Ku!. t; 



l>y Art. ;>0 these forc(\^ may be considered as acting at C. tlie 
point o( intiM'sccti(Ui o( ihc'w lines of action. Hence by the 
parallelogram law, a single force can be found, say R^ to replace 



66 



Elementary Mechanics 



the two forces P and Q. If any number of forces act, the paral- 
lelogram law applied first to two forces, then to their resultant 
and another force, and so on, will give a single force as a resul- 
tant, provided we are not required, in the course of this process, 
to combine two parallel forces. This case will now be considered. 

37. Parallel Forces. — We shall divide the consideration of 
parallel forces acting on a rigid body into three cases : 

Case I: When the forces act in the same direction. 

Case II: When the forces are unequal and act in opposite 
directions. 

Case III: When the forces are equal and act in opposite 
directions. 




Fig. 68. 



FoECES Acting on a Eigid Body 67 

38. Cases I and II.— Let P and Q (Fig. 68), be two parallel 
forces acting on a rigid body and let AB be any line cutting their 
lines of action in A and B, respectively. At A and B in the line 
AB put in two equal and opposite forces F^ and F^. These forces 
have no effect on the body. Let B^, the resultant of P and F^, 
and R^, the resultant of Q and F^, intersect at D. R^ and R^ 
may now be regarded as acting at D. We resolve R^^ and R^, 
acting at D, into components equal and parallel to their original 
components, i^^ and F^ annul each other and we have left P 
and Qj having the same line of action at D. 

In Case I, R = P + Q. 
In Case 11, R=P-Q. 

The resultant, R, may be regarded as acting at C, the inter- 
section of its line of action with the line AB. 
From similar triangles, we have 





AC 


_^i 












DC 


~ P' 












CB 


.F, 












DC~ 


~Q' 










S] 


ince Fj^- 


^F,, 


dividing. 


we 


have 




AC 
CB~~ 


P' 


or p. 


>AC: 


= Q 


• CB. 



Therefore, the resultant of two parallel forces is equal to their 
algebraic sum, is parallel to the forces, a^nd divides the line 
joining their points of application in the inverse ratio of the 
magnitudes of the forces. 

The line is divided internally when the forces act in the same 
direction, and externally when the forces act in opposite direc- 
tions. 

39. The Moment of the Resultant of Two Parallel Forces 
(except Case III). — From the theorem of Art. 35 the moment 



68 "Rl KMl-\ rVlxY "MlTHANU^S 

o( [he rosiilmnt of roiuurronl t\>i\H\^ is Ci\\]i\\ \o ilio sum of (l\o 
inonuMits o\' iho foroos about any lino \vli;Uo\or. Tlioroforo, ro- 
forrnii;- to I'li;-. (v*^. 

MouumU of A' -- nioniont oi' h\ | nuuuoiU o( l).. 

= inoiuoni o( /' | nuuuont o( /-', | niouionl oi Q | 

monuMit o( /■'.. 

= inoinoiu oi' /' I nioniont o( Q. 

'Vhcvc(ovc tho mouuMit o\' \\\c rosiillant o{ {mo pnrallol foroi>s 
about any lino is i\]ual to lhc> sum oi' tbo nuuuonts o\' tho two 
foroos about tbo lino. 

40, The Resultant of any Number of Parallel Forces. — l>y 

apply iui^- ropi\itodly tlu^ nuMlunl o( the proocnlini;- artiolos wo 
obtain tlio rosultant and momont o( any numbor o( j^aralK^l 
foroos. IKmu'o: 

(1) Tlio ri\<ul(ani ol' any numbor o\' parallol foroos is cH|ual 
to iboir ali^obraio sum aiul aots paralK^l io tbo foroos. 

(vM Tlu^ monuMit o( Ibo rt^sultant o\' any numluM- of parallol 
foroos with rospoot to any liiu^ is 0(]ual to tho sum o\' tho inomonis 
o( tho parallol foroos with rospoi't (o tho lino. 

Tluv^o iwo faois aro sullioiont to diMormino oomploli^ly iho 
rosultani, inohulin*;- mai;nit iult\ diiH^tion, ami lino o\' aotion. 

41. Coordinates of the Center of any Number of Parallel 
Forces. — In Arl. ;>S tho position o\' iho point (\ tho point o'i 
applioation ol' Iho iwsultant o\' two parallol foroi^s, aoliiii;- at .-t 
and />. is ind(^pond(Mit o\' i\)o dirootion o^ \ho fin'0(\s but d(^poiuls 
(Uilv upon iho points o\' applioaiimi and tlu^ ma;;niludo o( tho 
foroi\^. In liko mannor tlu^ iu>int ol' applioation ol' l\\o rosidtant 
ol' any numbor ol^ paralKd foroos, if tluMr poinls ol' applii-ation 
aro i^'ivcMi, is indopomKMil ol' tlu^ diriH'ticm ol' tlu^ fi>roc\s. 'Vho 
point is oallod Iho oonlor ol' tho parallol foroos. Us oo(U-dina(i\s 
may bo l\>und as follows: 



FoKci'S .\("i'iN(; ON A l\i(;ii) Body 



69 



lid. A',. /''.., I'\^ .... 1)0 llu' |);ir:ill('l I'dpccs, llu-ir points oC 
Hl>i)lir:ilioM (./•,, //,. :, ), (.r,. //,. :..), (.r.,. //.,, ;:,,) .... ( Ki«;. 
()!)). I><'i (•'■.//. -) 1h> Ihc (HMiltM- I'o llu> i)i)riilU'l I'orcrs. Since the 
ccnlcr is indcpnuUMil oi' i\\o ilii-cction of (he foi-ccs, \V(> shall 
assuino (lu> t'ori't>s paiallcl (o OZ. Taking nionirnl-; about 01', 
wo have 



A 



o ^1 
I 

.CL 






V\i\. GI). 



ii-{f'\ I F, I a:, . . . . ) /-Vr, I /'\.r, I A:,r, . . . . 



a;= 



Takini;- nionuMils alioui (^\'. 



i:/' 






l'\ I A', I A', .... 
r^v Icfiiui;- {\\c t"oi\'i>s act parallel to ( ^V and taking nioniouts 
about ()Z. 

-^ A',:, I A'.:., t A'.,:, , . . . _ :iFz, 

A\ I r, I a; .'. . . '~^F' 

}1S. l'\>rt>(^s o[' (? and 10 pounds have parallel lines of action 18 
inelu^s apart. Viwd the niagniluile id' their resultant anil the 
position ol' Its line o( action: (a) \vhen they have the same (.liree- 



70 



Elementary Mechanics 



tion; (b) when they have opposite directions. (In each case 
give the distance of R from the 10-pound force.) 

99. The lines of action of parallel forces, i^ = 2, F — ^, F — 1, 
F——4:, intersect a perpendicular plane in the following points: 
(1, -1), (0, 4), (-1, 3), (1, 2). What is their resultant and 
where does it act? 

100. Weights are hung on the rim of a wheel at the mid-point 
of each quadrant arc as follows: 1st quadrant 5 pounds, 2d 
quadrant 20 pounds, 3d quadrant 15 pounds, 4th quadrant 10 
pounds. What is the smallest additional weight that will keep 
the wheel from turning? 

101. The 16-foot beam AB is supported at P and Q (Fig. 70). 





. 




16' 


, 


B 


1 1 


A 


4' A 


\ ^ 


4' 


3' tn 


V4 




P 


> 


Q 


lOO 






v~ ^ 


\ 2C 


)0 - 





\x 



300 

Fig. 70. 



It weighs 210 pounds, and also carries the three loads, 200, 300, 
and 100 pounds, respectively, as per sketch. Find the weight 
that must be hung at A in order that the supporting force P 
may be 420 pounds. 

102. A table top is an equi- 
C lateral triangle with legs at A, B, 

and C (Fig: 71). 40 pounds 
rests at iZ" (5 feet from C), and 
eb pounds at ^ (4 feet from B). 
Find the total pressure on each 
leg. (Neglect weight of table 
top.) 

103. This same table top (di- 
mensions in feet) weighs 7 pounds 
per square foot and carries the 
same loads, but is supported by 

^ three legs (each in middle of a 
side). Find the pressure on each 
leg. 




Fig. 71. 



Forces Acting on a Eigid Body 



71 



104. Given a rectangle 6' x 8' and forces acting as shown in 
Fig. 73. Find moment of the forces about 0. 




42. Case III. Couples. — We shall now consider the effect on 
a rigid body of two forces which are equal and parallel and act in 
opposite directions. Two forces, equal in magnitude, parallel, 
and acting in opposite directions, but not at the same point, are 
called a couple. 

The distance between the lines of action of the forces is the 
arm of the couple. The plane determined by the lines of action 
of the two forces is the plane of the couple. 

If two equal and parallel forces act in opposite directions on a 
rigid body, they cannot be replaced by a single force, but their 
effect is not zero, for they have a tendency to turn the body 
around. In other words, their moment is not zero. 

43. Moment of a Couple. — The sum of the moments of the 
forces of a couple about any point of the body in the plane of the 
couple is constant and equal to the product of its arm and one of 
the forces. In Fig. 73 let P be the couple and a point in line 
with the arm of the couple. Taking moments about of the 
forces, we have 



72 Elementary Mechanics 

Momeut = P-OA-P.OB 

= P{OB + BA)-P.OB 
= P.BA. 

It can be shown that the effect on a rigid body is unchanged 
if one couple is replaced by another having the same moment and 
acting in the same plane or a parallel plane. The proof of this 




Fig. 73. 



is by repeated applications of the principles of Art. 30. It is of 
an elementary charact,er, but is omitted here on account of its 



length. 



44. Composition of Couples. — Theorem. — Any number of 
couples is equivalent to a single couple. Consider the plane M 
with the couples P and R (Fig. 74). Substitute for these two 
couples two other couples having the same arm c such that 



and 



P-a^S'C 

R'h=:T-C. 

Now at A there are actinoj two forces, the resultant of which is 



Forces Acting on a Eigid Body 



73 



the algebraic sum of the forces S and T. And at B the same 
resultant of forces acts. Hence the result of the substitution is 
a couple. Any number of couples may be treated in the same way. 




Fig. 74. 



45. A couple and a force acting in the same plane or parallel 
planes may be replaced by a single force whose line of action is 
determined as follows : 




Fig. 75. 



By Art. 43 we may replace the couple by a new couple which 
has one force acting along the line of action of the force P; then 
the resultant is equal to {P — Q)-i-Q = P, acting parallel to the 



74 Elementary Mechanics 

original force. Let be the point at which the resultant force 
acts. Taking moments around this point, we have 

or 

P'OA^Qi^AO^OB)^Q.AB. 

Therefore 0A = moment of couple divided by force P. Hence: 
The resultant of a couple and a force is a force w^iose magnitude 
is that of the original force, acting at a distance from the original 
force equal to the moment of the couple divided by the force. 

46. Theorem. — Any number of coplanar forces acting on a 
rigid body is equivalent to a force acting at any desired point in 
the plane and a couple. Let M be the point and F any one of the 




forces (Fig. 7G). Introduce two forces at M parallel and equal 
to F but acting in opposite directions. Then the force F has been 
replaced by a force F, acting at M, equal to the original force F 
and a couple. The same thing can be done with each of the 
other forces. The forces at M have a single resultant force acting 
at M, and the resultant of the couples is a single couple. (Art. 
44.) 

It can also be shown that any non-coplanar system of forces 
acting on a rigid body is equivalent to a single force acting at 
any desired point and a couple. 



Forces Actixg ox a Eigid Body 



75 



Prohlems 

105. Given the following couples, convert to equivalent couples 
by finding the forces corresponding to the indicated arms. 

Force 3 lbs. 6 lbs. 12 lbs. 14 lbs. 

Arm S f t. 7 ft. 12 ft. 6 ft. 

New arm 6 ft. 21 ft. 7 ft. 20 ft. 

106. Convert the foHowing couples to equivalent couples by 
finding the arms corresponding to the indicated forces. 

Force 5 lbs. 6 lbs. 10 lbs. 12 lbs. 

Arm 7 ft. 9 ft. 9 ft. 17 ft. 

New force S lbs. 9 lbs. 4 lbs, 5 lbs. 

lOT. Given two couples in the same plane, both having the 
same direction, namely 5-pound forces at 6 feet, and T-p"ound 
forces at S feet. Find the force of the resulting couple, if the 
arm is taken as 5 feet. 

lOS. Given these two couples in parallel planes and of opposite 
directions: 3 pounds at 9 feet, and 18 pounds at 6 feet, find the 
moment of the resulting couple. Also find the arm if the force 
is taken as 9 pounds. 



— 14— 



16lb5 



T 



P 5lbs 



Fig. 77. 



109. Given tlie force F and the couple P and F acting as in 
Fig. ;:. Find the magnitude, position, and direction of the single 
force which is the equivalent of the system. 



76 



Elementary Mechanics 



Review Problems 

110. A 60-pound cylindrical cask is 4 feet high and 3 feet in 
diameter. How long an iron rod weighing 
4 pounds per foot will just overturn the 
cask, placed as per sketch (Fig. 78) ? 

111. An iron rod weighs 10 ounces per 
inch length. It is lying on a table with 
3^ feet projecting over the edge, and a 
50-pound weight presses on the end lying 
on the table. How long is the rod if it is 
just about to turn? 

112. A rod 8 feet long and weighing 
40 pounds is supported at each end. How 
far from one end must a weight of 70 
pounds be hung in order that the support- 
ing force at that end be 50 pounds? 

113. The sides of a triangular table are 
3, 4, and 5 feet long. A weight of 24 pounds is placed on the 
table 8 inches from the 3-foot side and 18 inches from the 4-foot 
side. What is the thrust in each leg? (The legs are exactly at 
the comers. Neglect weight of table.) 





114. Given forces acting as in Fig. 79, also, ^0 = 15 feet, 
OB — S feet, and ON =12 feet. Find the single force, acting in 
the line HK which will have the same turning moment about 
the point as the four given forces acting together. 



Forces Acting on a Rigid Body 



77 



115. Given the force F, and the couple Q and Q, acting as in 
Fig. 80. Find the magnitude, position, and direction of the 
single force which is equivalent to this system. 



20lb$ 



71b. 



. 17'— ^8' 



Q 



^7lbs 



Fig. 80. 



116. A brakeman sets a brake by pulling 50 pounds with one 
hand and pushing 50 pounds with the other. His forces act 
tangentially to the brake wheel, whose diameter is 18". Another 
time he gets the same result by using a lever in the handwheel 
and pulling 25 pounds. How far from the rim of the handwheel 
must his hands be placed ? 

117. Parallel forces, each equal to P, act in the same direction 
at three corners of a square, perpendicular to its plane. At the 
other corner such a force acts that the whole system is a couple. 
Determine the moment of the couple if the side of the square is a. 



ts er^ 



78 



Elementary Mechanics 



CHAPTEE IV 

STATICS OF A RIGID BODY 

47. Equilibrium of a Rigid Body. — The resultant of a number 
of forces in a plane is either a single force or a couple. (Arts. 
45, 46.) Therefore, the condition for equilibrium is that the 
single force shall be zero and that there shall be no unbalanced 
couple. The component of the resultant force in any direction 
equals zero if the sum of the components of the several forces 
in that direction equals zero. That the resultant force may 
equal zero, its component parts in two different directions must 
equal zero. In this case there can be no resultant except a couple, 
and this will vanish when its moment about any point is zero. 
This can be expressed by the following equations; 

2Z = 0. 
SF=0. 
5-/¥ = 0. 

Thus a necessary and sufficient condition for equilibrium of a 

number of forces in a plane is : 

If the sums of the resolved forces 
in any two directions each equals 
zero and if the sum of the mo- 
ments about any point is also zero, 
the system of forces is in equilib- 
rium. 

48. Theorem. — If a rigid body 
is in equilibrium under three 
forces, these forces pass through a 
point or are parallel. 

Let A and B intersect at M 




Fig. 81. 



Statics of a Eigid Body 



79 



(Fig. 81) . The moments of A and B about M equal zero. Since 
the body is in equilibrium, the sum of the moments is zero. 
Therefore the moment of C about M is zero. Hence C must pass 
through M. 

If the two forces A and B are parallel, their resultant will be 
parallel to them ; and if the body is in equilibrium, the third force 
C must be equal and opposite to this resultant and have the same 
line of action. 

49. The foregoing principles are illustrated by the following 
examples : 

Example 1. — A uniform rod 12 feet long is supported by a 
smooth horizontal peg and the pressure of a smooth vertical wall 
against one end of the rod. The peg is 3 feet from the wall and 
the rod is in equilibrium. Find the angle 6 which the rod makes 
with the horizontal. 

Since the wall and the peg, P (Fig. 82), 
are smooth, the reaction, R, of the wall 
against the rod acts at right angles to the 
wall, and the reaction, S, of the peg 
against the rod acts at right angles to the 
rod. The weight, W, of the rod may be 
regarded as acting at the center of the rod. 
Taking moments about A, 

S.3sec^=lF. 



cos 0, 



Resolving vertically. 




COS^ 0: 



Fig. 82. 



^ = 37°2! 



Example 2, — A trap door 3 feet square weighing 50 pounds is 
held partly open by a cord 4 feet long. The cord is perpendicular 
to the plane of the door in this position and its ends are made 
fast to the middle of the door's outer edge and to a hook in a 
vertical plane through the hinges. Find the tension in the cord 
and the magnitude and direction of the thrust on the hinges. 



80 



Elementary Mecitaxics 




Since tlie door is in equilibrium under 
three forces, the reaction, R. pas>es throui^h 
(\ the intersection of T ami IV (Fig. 83). 
ABC is the force triaui^le. 







20 lbs. 



.\B 



irvi3 



= 10^'13 = 3G.1 lbs. 



tan $ = 



^y=«. 



t o-i . 



Pro ?)/('/??.<? 

118. A man weighing ISO pouiuls is half-way up a 20-foot 
ladder weighing SO pounds and resting against a smooth wall. 
The foot of the ladder rests upon a smooth floor, and is prevented 
from slipping by a horizontal cord attached to the wall. Find 
the tension of the cord, if the angle between the ladder and the 
floor is 30 ^ 

119. A uniform beam AB is 6 feet long, weighs 25 pounds, 
and rests at -1. upon a smooth floor, and at D upon the smooth 
top of the vertical post CD, 3 feet high. , A weight of 50 pounds 
hangs from B, and the end A is prevented from slipping by a cord 
AC, 4 feet long, connecting it with the foot of the post. Find the 
tension of the cord and the reaction of the floor at .1. 

120. A uniform rod 22 feet long, weighing SO pounds, rests 
with its upper end .1 against a smooth vertical wall, and its lower 
end supported by a cord, ^i) feet long, attached to a point directly 
over .1. Find the tension of the cord, and the resistance of the 
wall. 

121. A trap door 3 feet square is held at an inclination of 30'' 
to (and above) the horizontal plane through its hinges by a cord 
attached to the middle of the side opposite the hinges. The other 
end of the cord, which is 5 feet long, is attached to a hook verti- 
cally above the middle point of the hinged side of the door. Find 



Statics of a Eigid Body 



81 




the tension in the cord, and the direction and magnitude of the 

pressure between the door and its hinges, 

the weight of the door being 50 pounds, ^c 

which may be taken as acting at the center 

of the door. 

122. The supporting cross bar, AB, of 
a platform is 6 feet long and holds weights 
as shown (Fig. 84). The inner end is 
fastened to the wall by a hinge at ^. A 
cable BC at an angle of 30° with the ver- 
tical supports the other end. Solve for 
the tension in the cable BC and the reac- 
tion at A in amount and direction. 

123. A uniform rod 10 feet long weigh- 
ing 20 pounds rests with one end at the 
junction of a vertical wall and a horizontal 
floor; from a point 2.5 feet from the other 
end a cord runs horizontally to a point 

6 feet above the floor. Find the tension of the cord and the 
amount and direction of the reaction at the lower end of the rod. 

50. Two-Force Pieces. — Many structures, such as roof and 
bridge trusses, are made up of members connected by pins, and, 
neglecting the weight of the members themselves, each member 
is acted upon by only two forces, acting at the joints. Since each 
member is in equilibrium, these two forces must be equal and 
opposite in direction, that is, the line of action of the forces is 
the axis of the member. These are called two-force pieces. If 
a member is in tension it is called a tie, and if it is in compression 
it is called a strut. 



800 

Fig. 84. 



51. Method of Sections. — The method of sections is a method 
used to determine the internal stresses in the members of the 
structure. It consists in cuttino- a section throuojh a structure 
and considering the part on one side of the section as a rigid body, 
held in equilibrium by the external forces acting on this side of 



82 



Elementary Mechanics 



the section and the stresses along the members cut by the section. 
To explain more fully, we will find some stresses by this method. 




Fig. 85. 



Consider the forces acting as illustrated by the figure (Fig. 85). 
First find the reactions at A and E. Taking moments of all the 
external forces about A, we have 



Therefore 



§.40 = 217(10 + 20 + 30), 
Q = SW. 

P = 3W. 



Let it be required to find the stress in BG. In Fig. 86 let 




Statics of a Eigid Body 83 

F = 2000 pounds. Take a section cutting BC, BG, and HG, as 
shown. Indicate the forces as acting away from the section. 

To get S2 resolve vertically (SF = 0). 

6000-4-000-/52 cos 45° =0. 
Therefore 

6'2 = 2000V2lbs. 

Since the result is positive, the stress acts away from the joint 
B, and therefore is tension. If the sign had been negative, the 
stress would have been toward B and therefore compression. 

We can find S^ by taking moments about B, the intersection 
of Sj^ and S^. 

6000. lO-S^' lOrrO; 
/S'3 = 6000 lbs. 

The positive sign indicates tension as before. To find S-^, resolve 
horizontally (:§Z = 0). 

>S'3 + ^2COs45°+/Si = 0; 
6000 + 2000 ^-iSi^O; 
>Si= -8000 lbs. 

The negative sign indicates that the stress acts toward B and is 
therefore compression. 

If we had wished to compute only one force, say S^, the 
simplest method would usually be by taking moments about the 
intersection of the other two unknown stresses. 



Thus 



S^ ' 10 + 6000 . 20-4000 • 10 = 0; 
>S',=:-8000; 

= 8000 lbs. compression, as above. 



84 



Elementary Mechanics 



Problems 
134. In the truss of sketch (Fig. 87) the peak carries a con- 
centrated load of 800 pounds, and in 
addition to that the left rafter car- 
ries 60 pounds per foot run. Find 
the stress in the loaded rafter by 
means of a section. 

125. The truss of sketch (Fig. 
88) carries 200 pounds per foot run 
over PAQ. Find the amount and 
kind of stress in the vertical mem- 




FiG. 87. 



ber by means of a section. 



15 



15 



Q 




Fig. 88. 

126. The truss of sketch (Fig. 89) carries two concentrated 
loads, as shown. Find by the method of sections the amount and 
kind of stress in the member marked (*). 




Fig. 89. 



lOOIbs 




Fig. 90. 



Statics of a Eigid Body 



85 



127. In the truss of sketch (Fig. 90) there are two concen- 
trated loads carried, as shown. Find by the method of sections 
the amount and kind of stress in the member between the loads. 

128. The girder of the sketch (Fig. 91) carries three concen- 
trated loads, as shown. Find the amount and kind of stress in 
the two members marked (*), using the method of sections. 





20' 


26 


20' 


20' 




zd 


\ 




y 




2d 






zd 


, 




w 



o 


o 





o 

















CO 


(M 


CO 



Fig. 91. 



129. Given the cantilever conveyor of the sketch (Fig. 92). 
Find the amount and kind of stress in the two members marked 
(*) when the frame is loaded as shown. 




Fig. 92. 



52. Maxwell's Method. — The graphic method, invented by 
Clerk Maxwell, is based upon the fact that since the forces are in 
equilibrium they may be represented in amount and direction by 
the consecutive sides of a closed polygon. If the loads and 



86 



Elemextary Mechaxics 



external reactions are all vertical, the polygon of the external 
forces reduces to a vertical straight line. 

Let the loads be as indicated in the frame diagram (Fig. 93). 




RJ3T 




B 


Aa=.Dd=3\'2 [O 




Bo:=zCc^2\ 2 [O 


E 


ab = cd=\2 [C) 


C 


Ea = Ed = 3 (J) 




hc=:2 (T) 



Fig. 94. 

To name these forces, place a capital letter between successive 
external loads and reactions and a small letter in each space made 
by the members of the frame. In this way each load and each 
member is straddled by a pair of letters. Each force is designated 
by the letters on each side of it. 

To draw the reciprocal diagram (E. D.) . first lay off to scale the 
external forces and reactions in order. Thus the loads between A 
and B. B and T. T and D. and the reactions between D and E. and 
E and A, in the frame diagram, are represented in amount and 



Statics of a Eigid Body 87 

direction by the lines AB, BC, CD, DE, and EA, in the recipro- 
cal diagram (Fig. 94:), forming a closed line. To represent the in- 
ternal stresses in the members of the frame on the R. D., we draw 
lines parallel to the members of the frame. All the forces acting at 
a joint in the frame form in the R. D. a closed polygon. Consider 
the forces acting at the right support Ro- ^^ ^2i\Q DE — R^, 
already drawn. At E in the R. D. draw a line Ed parallel to the 
horizontal member Ed. Then d will lie somewhere on the line 
Ed. From D in the R. D. draw Dd parallel to Dd of the frame. 
d will lie at the intersection of these two lines. The triangle DEd 
is thus the force polygon for the three forces acting at R^. Next 
draw the force polygon for the joint i^i- Draw Ea of the R. D. 
parallel to Ea of the frame and from A draw Aa parallel to Aa 
of the frame, a will lie at their intersection and will coincide 
with d. To determine c draw from C a line Cc parallel to Cc of 
the frame, and from d draw dc parallel to dc of the frame. Their 
intersection determines c. In like manner the intersection of Bh 
and ab determines t. The line joining & and c will be vertical 
and parallel to he oi the frame, and checks the accuracy of the 
drawing. 

To determine the Icind of stress in a member of the frame, 
let "Qs consider again the forces acting at Ro- These forces will 
form a closed polygon, viz., the triangle 
DEd in the R. D. (Fig. 95) . The sides of 
the triangle represent in amount and direc- 
tion the forces acting at R. Since we know 
DE acts upward, Ed acts to the left, and 
dD acts downward, as indicated by the 
arrows. Since Ed acts to the left, it acts 
away from the joint, and therefore Ed is pjQ_ 95_ 

in tension. Since dD acts downward, the 
stress is toward the joint, and therefore dD is in compression. 




88 



Elementary Mechanics 



Consider the stresses at the joint between C and D. These 
are represented by the polygon CDdc (Fig. 9G) in the K. D. We 
know that CD acts downward. This gives 
the forces and directions as indicated by the 
arrows. It should be noted tliat the arrow d< 
on Del here points upward, while in Fig. 95 
it points downward. But here we are con- 
sidering the forces at tlie joint between C 
and D, and the force is indicated as acting 
toward the joint, showing compression. In 
the former figure we were considering the 
forces at the point of support; the stress in 
Dd acted toward the su])])0)-t, and indicates compression in both 
cases. 




Problems 
130. The rafters of a simple triangular roof truss are 20 feet 
long, and the span is 33 feet (Fig. 97), 



Find the amount aiul 



300lb 



600 lbs 




300lh 



Fig. 97. 



kind of stress in each member when loaded with GOO pounds at 
the peak. Use the R. D., aud find the stress in the left-hand 
rafter (as a check) by (he method of sections. 



Statics of a Rio in Body 



89 



K>1. The roof truss of skotoli (Fii^. tV^) lias a vtM-lii'al iiiombor, 
and carries 'vH)00 })Oiuuis at tho peak. Kind anunmt and kind ol: 
stress in the two members marked (*). 



:ooOiioi 




1000 



1000 



Fio. 9S. 




132. Find the anunmt and kind of stress in the members 
marked (*) in tlie live-member truss of sketch (Fig. 99). The 
joints E and F carry concentrated loads of 800 pounds and GOO 
pounds respectively, and the member PE carries a uniform load 
of 100 pounds per foot run. 



2240 




133. The sketch shows a roof truss with a vertical member 
(Fig. 100). The load on the rafters is TO pounds per foot, and 
the load on the horizontal members (due to lloorii\g) is 20 poimds 
per foot. Find the anunmt and kind of stress inthe four mem- 
bers nuirked (*), and check the stress in the vertical nunnber by 



taking a section. 



90 



Elementary Mechanics 



131. This roof truss {Fig. 101) carries 31 tons at tlie peak, 

21 T 




Fin. 101. 



and a imilorni load of ^ ton pci- Foot run over PQ. Find amount 
and kind of stress in all five members. 

135. A truss with 8-foot vertical member is shown in sketch 
(Fig. 102). The 17' rafter carries 200 pounds per foot, and the 




Fig. 102. 



10' rafter 80 pounds per foot. Also 4200 pounds acts at joint 
Z. Find amount and kind of stress in members marked (*). 

53. Three-Force Pieces. — Ilci-ctofore wo have considered cases 
where members of a structure are acted upon by two forces only 
and the members have been subject to either compression or 
tension. If a member is a(;tcd upoii by three or nu)re forces, 
usually none of tliem will act alon^- tlie axis of tbe member. Con- 
sider the following figure to illustrate the problem : 

Find the tension in the rope BC and the amount and direction 
of the reaction at A, if 17 = 2800 pouiuls (Fig. 103). The rod 



Statics of a Eigid Body 



91 



AB is acted upon by three forces : The tension T in EC, acting 
in direction BC ; the force W, actin^- downward at D, and re- 



zM^ 




Fig. 103. 



Fig. 104. 



action at A, unknown in amount and direction. We will resolve 
this reaction into its horizontal and vertical components, Bu and 
By, as shown in the Fig. 104. 

To find the tension T, take moments about A. 

T . 11- sin 15° = 2S00 . 10 cos V^° ; 
r=:51Gl.llbs. 
Resolving horizontal!}^, (2A'=:0), 

TcosGO°+i?//=0 .\Bu=-2^o21 lbs. 
Resolving vertically, (51'=: 0), 

Tcos 30°-2S00 + /m- =0 .*. 7?,-= -1932.1 lbs. 

The negative signs of 7?7i and 7? r indicate that the actual direc- 
tions of these forces are 0}>posite from what is assumed in the 
ligure. 



^mtan 



1932.1 



-2T32.1 
A^ = i?//sec (9 = 3316.2 lbs. 



= 35nG' + lS0^^=215°16'. 



92 



Elementary Mechanics 



Problems 

136. Find the stresses and reactions in the frame of sketch, 
Fig. 105, loaded as shown. Neglect the weights of the members. 




8- -^-4- 



3300lbs. 




E M/ 



lOT 



Fig. 105. 



Fig. 106. 



137. Find the stresses and reactions in the framework of the 
sketch, loaded as shown (Fig. 106). Neglect weights of the 
members. 




Fig. 107. 



Fig. 108. 



138. In the crane shown in Fig. 107, find the forces acting on 
the pins and the tension in the tie AG. 



Statics of a Eigid Body 



93 



139. Find the amo-ant and direction of the pressure of the pin 
at B on the member ABC, of the pin C on the member ABC, 
and of the pin at D on the member BD (Fig. 108). 

140. A uniform spanker boom weighing 400 pounds and 32 
feet long is topped up to 30° with the 

horizontal by the topping lift which is 
fast to the boom's end, and in thisC 
position makes an angle of 60° with 
the boom (Fig. 109). Find the ten- 
sion of the lift, and the amount and 
direction of the thrust at the goose- 
neck. 

141. The upper end of a 150-pound 
beam rests against a smooth vertical 
wall, 3 feet above the floor. The lower 
end is fastened to a smooth hinge in 
the floor 8 feet from the wall. Find 
the wall resistance, the hinge resistance, ^ 
and the angle the latter makes with 
the floor. 




Fig. 109. 



Revieiu Problems 

142. In the truss of sketch (Fig. 110) the rafters carry 30 
pounds per foot run and meet at right angles. Find the amount 
and kind of stress in the member marked (*) by means of a 
section. 




Fig. 110. 



94 



Elementary Mechanics 



143. Find reactions R^ and R^, and the stresses in the different 
members of the truss of sketch (Fig. 111). 



16 T 



16 T 




144. Points of application of forces are equidistant, and K is 
on the perpendicular bisector of AD (Fig. 112). Find (1) the 
magnitude and direction of reaction at G, the joint A being 




Fig. 112. 

mounted on frictionless rollers; (2) the magnitude and direction 
of reaction at A. 



12 T 




Fig. 113. 



Statics of a Rigid Body 



95 



145. Determine stresses Bb, At, ah, Aa, Ca, and determine W 
if the stay Aa makes an angle 30° with the vertical. Determine the 
reaction at 5C (Fig. 113). 

146. The truss sketched (Fig. 114) carries 40 pounds per 
foot on the rafters, and 20 pounds per foot on the platform PQ 




Fig. 114. 



(due to flooring). Find the amount and kind of stress in the 
four members marked (*), using the R. D. Determine the kind 
of stress in the rafters and the vertical member by drawing a 
separate sketch of the polygon of forces for the peak joint, as it 
appears in the R. D. Also, find the stress in the right-hand 
horizontal member by the method of sections. 



30T 30T 30T 




147. Draw the R. D. for the truss sketched (Fig. 115), and 
find amount and kind of stress in the members marked (*). 
Then find the stress in the left-hand lower horizontal member 
by means of a section. 

148. In the Warren girder of sketch (Fig. 116), PQ carries a 
load of 50 pounds per foot and also has a concentrated load of 
1200 pounds at its center (mid point of PQ). Each member is 
10 feet long. Draw the R. D. and find amount and kind of stress 
in the members marked ( * ) . 



96 



Elementary Mechanics 
10 10' 




Fig. 116. 



149. The Queen Truss of sketch (Fig-. 117) carries two loads 
(9 and 18 tons), as shown. Find anioimt and kind of stress in 
the members marked (*). 





9T 


,s 


T 




M 


\ 6' 


8' b ^..^ 
^^^^ d 


6 

V 


m 



15 
D 

Fig. 11' 



150. Given the Queen truss of sketch, loaded over the struts 
as shown (Fig. 118). Draw the E. D. and compute the stresses 



15T 



6 T 



A 
6 > 


B 
15 ., 




C 

. 6 


«\e 


^.^ d 








Fig. lis. 







in the four members marked (*), and state the kind of stress in 
those members. 

151. The truss of sketch (Fig. 119) carries a concentrated 
load of 2200 pounds at the peak, also the 15-foot member carries 



Statics of a Eigid Body 



97 



a load of 80 pounds per foot. Draw the E. D., and find amount 
and kind of stress in the members marked (*). 



2200lbs 




152. The A-frame shown in Fig. 120 supports a load of 10,000 
pounds at the middle of a member BD, which is 18 inches from 




Fig. 120. 



the floor. Determine the pin reactions at B, C, and D caused by 
this load, if the floor is considered to be smooth. 

153. Consider each member of the A-frame in Ex. 152 to weigh 
100 pounds per linear foot and determine the pin reactions at B, 
C, and D. 

154. K pentagon ABODE, formed of equal uniform heavy 
rods, connected by smooth joints at their ends, is supported sym- 
metrically in a vertical plane with A uppermost, and AB, AE, 
in contact with two smooth pegs in the same horizontal line. 
Prove that if the pentagon is regular the pegs must divide AB 
and AE each in the ratio 



- sinY^Q-TT 



3 sin J 



\- 



98 Elementary Mecpianics 



CHAPTER V 
CENTER OF GRAVITY AND MOMENT OF INERTIA 

54. Gravitational Forces. — The formulae of Art, 41 determine 
a point on the line of action of the resultant of a system of paral- 
lel forces. The forces due to gravity, the pull which the earth 
exerts on bodies situated on its surface, we may regard without 
appreciable error as constituting such a system of parallel forces, 
since the dimensions of any body we shall study are very small 
when compared with the radius of the earth. Such a system of 
parallel gravitational forces acting on a body will have a resultant 
whose line of action, no matter what the position of the body in 
space, will always pass through the same point in the body; for 
it was shown (Art. 41) that the point of application of the resul- 
tant of a system of parallel forces does not depend on the angle 
which the lines of action of the forces make with the coordinate 
planes, but that it is completely determined by the magnitudes of 
the forces and by their points of application. The moment of 
the resultant force about this point through which its line of 
action passes is obviously zero. This leads us to the following 
definition. 

55. Center of Gravity. — The center of gravity of a body is a 
point G, such that if the body be supported by any axis what- 
ever through G, it will have no tendency to turn about that axis. 
That is, about any axis through G the sum of the moments of the 
gravitational forces acting on the particles of which the body is 
constituted must be zero. 

56. Coordinates of the Center of Gravity. — In order to deter- 
mine in a given body this point G, we choose, to suit our con- 



Center of Gravity 



99 



venience, three mutually perpendicular planes, constituting the 
planes of reference for a system of Cartesian coordinates, and 
regard them as rigidly attached to the body. As the body is 
moved about in space, these planes move with it and the values of 
the coordinates x, y, z, which indicate the distances of any point 
in the body from these planes, are in no way affected by its 
change in position. In order to determine the three coordinates 
{x, y,'z) of the point G we take moments about each of the three 
axes through G, parallel respectively to the X-, Y-, and -Z'-axes of 



2 


— 


^ 


1 


7r~^Ax.Lj,z) 


\ / 


/ / y 


7 / 


/ 




/ 


V/ 


_^ 



Fig. 121. 



the system of coordinates which we have just set up. Place the 
body with the X- and F-axes horizontal and the Z-axis, therefore, 
parallel to the lines of action of the gravitational forces, and con- 
sider the tendency of these forces to produce rotation about an 
axis through G, parallel to the F-axis (see Fig. 121). We first 
study the moment due to a piece of the body of mass Am, of 
any convenient size and shape, subject only to the restrictions 
that Am shall lie between two planes, each parallel to the YZ- 
plane and distant respectively x and x-^Hx from it, and that Am 
shall approach zero as Lx approaches zero. Denote the resultant 



100 Elementary Mechanics 

of the gravitational forces acting on Am by AW. Then AW = 
JcAm, where ^ is a constant for all bodies at the same place on the 
earth's surface. (Tlie value of k also depends upon the units of 
force and mass employed, see Art. 122.) The distance of the line 
of action of AW from the Z-axis is between x and x-\-Ax. There- 
fore, taking moments about an axis through G parallel to OY, 
the arm is x-\-e- Ax — x and 

Moment oi AW = {x — x) -k - Am + ek • Ax • Am, 

where e is less than 1. The sum of the moments of all the gravi- 
tational forces acting on the body is given by 

Ax='o 2_i [ (^ ~ ^') -^ ' ^^ + €^' • ^^ • ^^] y 

where the summation is to cover the entire mass. By theorems 
of the Calculus this sum is equal to the definite integral of {x — 
x) -k • dm, taken between the proper limits so that the integration 
is extended over the entire mass of the body. Whether this calls 
for single, double, or triple integration, we shall denote it by 

(x — x) 'k • dm. 

M 

From the definition of the center of gravity (Art. 55) the sum 
of the moments represented by this integral is zero. Hence 



(1) \{x-x). 



dm — 0. 

M 

Proceeding in a similar manner, taking moments about an axis 
through G parallel to the Z-axis, we obtain 



(2) 



{y~y) 'k'din — O. 



(4) 



Center of Gravity 101 

Placing the body so that the X- and Z-axes are horizontal and 
proceeding as before, we get 

(3) [(^-;s) .^^.^m = 0, 

M 

Since x, y, and z, and li are constants for the integration, we can 
solve the equations (1), (2), (3) and obtain 

X dm ij dm z dm 

dm dm dm 

M M M 

The denominator dm is the mass M of the body. 

M 

57. In future applications of the summation theorem of the 
Calculus we shall not indicate the several steps of the procedure 
as in the last paragraph, but describe it in some such manner as 
this : Take an element of mass dm at a distance x from the YZ- 
plane. The moment of the gravitational force acting on it is 

Ic-X' dm. Integrating to cover the entire mass, we have U; • ^ • dm, 

M 

or the sum of the moments of all the gravitational forces acting 
on the body. 

The derivation of equations (4) shows that the size and shape 
of dm are not fixed. We choose an infinitesimal element of 
length area, or volume — ds, dA, or dV respectively — as the prob- 
lem in hand may require, precisely in the way already familiar 
in the corresponding problems of finding length of arc, area, 
or volume by methods of the Calculus. We are subject only 
to the additional restriction that in computing x, y, z, all points 
of this infinitesimal element shall in the limit be equally distant 
respectively from the YZ, XZ, or XY planes. Then dm is 



102 Elementary Mechanics 

equal to ds, dA, or dV, as the case may be, multiplied by the 
density factor p, where p is the mass of a unit volume. 

58. If we write the first of equations (4) as \ x-k - dm^x-h -M, 

M 

the left member gives the sum of the moments about the F-axis 
of all the gravitational forces acting on the body of mass M ; the 
right member is the moment of the resultant of all these forces 
acting at the point G. Since we may take any line as the F-axis, 
this shows that about any axis the moment of the gravitational 
forces acting on the body can be replaced by the moment of the 
force due to a particle endowed with a mass equal to that of the 
given body and situated at its center of gravity. 

59. Center of Mass; Centroid. — It is apparent from equations 
(4) that the point G is determined by the distribution of the 
mass of a body quite independently of the action of the gravita- 
tional forces, and that, consequently, the point G might with 
equal correctness be called the center of mass of the body. 
Further, if the mass be uniformly distributed through the space 
occupied by the body, that is, if p be constant, the density factor p 
can be cancelled from the numerator and denominator in equa- 
tions (4), thus giving x, y,z a purely geometric definition. The 
center of gravity of a purely geometric quantity, arc, area or 
volume, is sometimes called its centroid. Any plane, axis, or 
point, of symmetry contains the centroid, a fact which should be 
made use of whenever possible to avoid unnecessary computation. 

60. Illustrative Examples. 

Some of the further simplifications that can be made use of 
in applying these formulae will be shown in the examples that 
follow. In particular, whenever the value of the denominator is 
a well-known result from the Geometry or the Calculus, only 
the numerator will need actually to be integrated. 



Center of Gravity 



103 



'2,4) 



(2^) 



Example 1. — Find the center of gravity of a plate weighing 
w pounds per square foot, bounded by the parabola y^ — Sx and 
the double ordinate for which x — 2 feet. 

Solution. — The graph of one face of the plate is shown in 
Fig. 122. If we consider the thickness of the 
plate it is evident from considerations of sym- 
metry that the center of gravity will lie in the 
plane equidistant from the faces of the plate, 
but it is customary in the case of thin plates to 
neglect the thickness and proceed as if we had 
to consider an area only. Then z = 0, and since 
the X-axis is an axis of symmetry for the area 
we have also ^ = 0. To find x we take as the 
element of area a strip dx wide parallel to the 
F-axis, whence dA = 2y dx. Since p — w, we 
have 

dm — pdA = 2yiv dx. 

Substituting in the first of formulae (4), 

J ^^^ f ' 2wxy dx 2 V 2 f ' x^ dx 

-_M __lo Jo 

{dm l2wydx 2V2\'x^dx 



\: 



(2.-4) 
Fig. 122. 



Therefore the center of gravity is on the axis of the plate, f feet 
from its vertex. 

Example 2. — Find the center of gravity of a rod I feet long, 
whose density varies as the distance from one end, the weight 
increasing from pounds per foot run to w pounds per foot run. 

Solution. — Since the only variable dimension that enters into 

n Liij 1 



Fig. 123. 



this problem is the distance from one end of the rod, we may 
regard it as a straight line having mass. Then 

p — x/l . w, dm = x/l ' w • dx. 



104 



Elementary Mechanics 



whence, cancelling a constant factor common to numerator and 
denominator. 



X dm x^ dx 



V'" i 



X dx 



= 2V3. 



Example 3, — Find the center of gravity of a right circular 
cone when the density of the cone varies as the distance from a 
plane through the vertex, and parallel to the base. 

Solution. — Suppose the cone to have altitude Ji and base radius 
a. Take the axis of the cone as the Z-axis, the origin at the ver- 




FiG. 124. 



tex of the cone. From symmetry t/^O, i = 0. To find x, take 
as the element of volume dV , a cylindrical slice dx thick, with 
bases parallel to the FZ-plane. 

Then 

p — liX, dV = 7rz^dx, 
whence 

dm = pdV='k7rxz^dx. 

From Fig. 124 z/x = a/h. Then 



Center of Gravity 



105 



A 



X dm 



dm 



[h 
1cit\ X'Z" 
— Jo 

Jo 



dx 



xz^dx 



x'^dx 
x-'dx 



4/i/5. 



The student should note that the method of this example is appli- 
cable to finding the center of gravity of any solid of revolution. 

Example Jf. — Find the center of gravity of the area bounded 
by the cardioid r = a ( 1 + cos (/>), and the initial line OX. 

Solution. — Take an element of area with dimensions dr by 
T'd^ and integrate first with respect to r to include the sector 
OAB (Fig. 125), and then with respect to <^ from to tt to sum 
all such sectors. 

dA=r-d<f)- dr, 
and since 



a;=r cos <\>, 
we have 

\x dm 



— r sin <^, 



TT fa(l + cos 0) 



r cos 



dr 



y 



\dm 
I dm 



\ area cardioid 

(l + COS 0) 

r- sin (^ dr d4> 




area cardioid 



Fig. 125. 



The denominator has the value 

fir fa(l + cos 0) 
r'dr.dcf> = 37ra^/4.. 
Jo 

The evaluation of the above integrals gives the results 
x = 5a/6, y = 8a/97r. 

Where more than one of the coordinates of the center of gravity 
is to be determined by integration, care should be taken to 
compute the denominator integral intact, without the cancel- 
lation of any numerical factor. 

Example 5. — Find the center of gravity of the area of one 
quadrant of the ellipse x = a cos cf), y = h sin </>. 



106 



Elementary Mechanics 



• Solution. — Consider the first quadrant, described as <^ varies 
from to 7r/2 (Fig. 126). Hav- 
ing cancelled the constant density 
factor p, the denominator integral 
is one-fonrth the area of the 
ellipse, i. e., Trab/4,. We shall set 
np the numerator integral for x 
in two ways, first by taking for 
dA a slice parallel to the F-axis, 
dx wide and y long; second by 
taking for dA a slice parallel to 
the X-axis, dy wide and x long. 
In the first case all elements dA 
are equally distant from the Y- 
axis, and we have directlv 




Fig. 126. 



Jo: JJ. ^J xydx=-Y^(^aco^<l>) {b sin <f>) ( 



a sin </) d<f)) 



2 a^h sin- </> cos </> 



■ a-h/3. 



li, however, we take a strip dA — x-dy, we violate the restriction 
to which we have previously held, that all points of dm should, 
in the limit, be equally distant from the axis of rotation. But 
we have seen (Art. 58) that the moment about OF of the strip 
x-dy is equal to the moment about OY of an equal mass con- 
centrated at the center of gravity of the strip, namely a:/2 units 
from F-axis. Hence the moment of the element d A— x-dy is 
{x- dy)x/2, and the moment of the entire area, which is the 
numerator integral, is obtained by summing this to include the 



entire area. 



Therefore, 



Hence in place of he • dA, we have 

M 

f f- 

I {x-dy)/2 = a-h/2reos^<j>dcl> = a~h/3. 



x= (a~'b/3)/{7rah/4.) 
and a similar computation will show 
y = 4:h/37r. 



4a/37r, 



Center of Gravity 107 

Prohlems 

155. Find the distance from the base to the center of gravity 
of a triangle with altitude h and base 6. 

156. Find the distance from the center of a circle with radius a, 
to the center of gravity of a sector of angle 2 A. 

157. Find the center of gravity of a right circular cone with 
altitude li and base radius a. 

158. Find the center of gravity of a hemisphere of radius a. 

159. Find the center of gravity of the area bounded by the X- 
axis and one arch of the cycloid x = a(c^ — sin<^), y — a{l — 
cos (^). 

160. Find the center of gravity of a circular arc of radius a 
and subtended by a central angle 2 A. 

161. Find the center of gravity of the lateral area of a cone 
of revolution with altitude /i and base radius &. 

162. Find the center of gravity of the frustum of a para- 
boloid of revolution having a single base of radius &, the height 
of the frustum being li. 

163. Find the center of gravity of the arc of the half arch of 
the cycloid a: = a(<^ — sin (/>), ^ = ^(1 — cos c/)), included between 
(/) = and </) = 7r. 

164. Find the center of gravity of the area between the curves 
y^ = 8x and x- = 8y. 

165. Find the center of gravity of a hemisphere when the 
density varies inversely as the square of the distance from the 
center. (Use a ring-shaped element of volume with cross section 
r-dcf)- dr.) 

166. Find the center of gravity of the arc of one quadrant of 
the hypocycloid x^-\-y^ = a^. 

167. Find the center of gravity of the solid formed by revolv- 
ing the cardioid r = a( 1 + cos (/>) about its axis. 

168. Find the center of gravity of the area between the cissoid 
y^(a — x)—x^ and its asymptote. 

169. Find the center of gravity of any zone of a sphere. 

170. Find the center of gravity of a, circle of radius a if the 
density varies as the nth power of the distance from a given 
point on the circumference. (Use polar coordinates.) 



108 Elementary Mechanics 

171. Find the center of gravity of one arch of the curve 
y = smx; of the area between this arch and the X-axis; of the 
surface and volume generated by revolving it about the X-axis. 

172. A right circular cylinder, base radius a/ axis OZ, is cut by 
a plane making an angle of 45° with its base and intersecting the 
base in OY. Find the center of gravity of that part of the sur- 
face and of the volume bounded by this plane, the XY and XZ 
planes. 

173. Find the center of gravity of the volume obtained by 
rotating^ about the X-axis the first quadrant of the ellipse h^x^-{- 

174. Find the center of gravity of the surface generated by 
revolving the cardioid 7^=a(l + cos </>) about its axis, 

61. The Center of Gravity of Composite Bodies.^ — The equa- 
tions (4) , Art. 56, may be written in the form xdm = x . M. Tlie 

M 

integration indicated in the left-hand member of the equation 
must be carried out to cover the entire body. In many problems 
we know the results of the integration for the parts into which 
the body may be divided. The value of the integral is, then, sim- 
ply the sum of these results and consequently we may find the 
values of x, y, z without actually performing the integration. 
Thus if a body of mass M is made up of pieces having masses 
M^, if 2, .... Mn and centroids {x-^, y^, z-^), {x^, y^, z<>), . . . . 
{xn, yn, Zn) respectively, and the integration for the first part 
gives M^x^, for the second part M^x^, etc., then the coordinates 
of the center of gravity of M, {x, y, z), must satisfy the relations 

xM = xJ\I^-{-X.J^L^J^. . . . +XnMn, 

yM = y^M^ + y^M^^. . . . +ynMn, 

ZM = Z^M^ + Z^M,+ . . . . +ZnMn, 

The following examples illustrate this procedure. 

Example 1. — Find the centroid of the cross-sectional area of 
the steel beam shown in Fiff. 127. 



Center of Gravity 



109 



6- 



C 



,B 



6" 
Fig. 127. 



Solution. — For convenience take the axes as shown in the 
figure. Then x-=0 from s}Tiimetry. 
About any axis, moment of total area = 
moment 45 + moment CD -{-moment ! 6" 

EF, where AB denotes the area of the 
rectangle of which A and B are oppo- 
site vertices, etc. The centers of grav- E 2" 
ity of these rectangles have ordinates 
1, 5 and 9.5 respectively. Hence, taking 
moments about the X-axis, we have 

y/(12 + 12 + 18)=12xl + 

12x5 + 18x(9.5), 

whence y/ = 243/42 ==5.8 inches from the 

bottom of the beam. 1 . 'v^ ^^, 1 — x 

Equally well we may obtain the 
result by using the equation : 
Moment of given area = moment AF — 
moment BD — moment CE. Then 

• 7/(66-12-12) =66 X (5.5) -12x5 -12x5, 

and therefore, 

^ = 243/42 = 5.8. 

Both these solutions can be obtained from the single formula 

y M^ + M^ + M^ 

if we agree that any mass added shall be given a positive sign, 
and that any mass cut out shall be given a negative sign. 

Example 2. — Find the center of gTavity of a circular area of 
radius 8 inches in which a hole of radius 2 inches has been cut, 
the center of the small circle being 5 inches from the center of the 
large one. 

Solution. — About any axis, the moment of the large circle = 
moment of final area + moment of small circle. Take the line of 
centers as the X-axis, the origin at the center of the large circle, 
as indicated in Fig. 128. 



110 



Elfmentary Mechanics 




Fig. 12J 



Taking moments about OYj 

647r.0=(647r-47r) •X + 4:it-5, 

whence 

x=-i. 

Or directly from the formula 

_ 647r.0-47r.5 

a;= — 

64-77 — 47r 

From symmetry 



X=: 



— _1 



Note. 
tion. 



Prohlcms 
-These problems are to be solved without the use of integra- 



175. Find the center of gravity of the trapezoid whose vertices 
have the coordinates (0,0), (12,0), (12,16), and (0,10). 

176. A uniform plate of metal 10 inches square, has a hole of 
3 square inches cut out of it, the center of the hole being 5/2 
inches from the center of the plate ; find the position of the center 
of gravity of the rest of the plate. 

177. Where must a circular hole 1 foot radius be punched 
from a circular disk of 3 feet radius so that the center of gravity 
of the remainder may be 2 inches from the center of the disk ? 

178. The mass of the moon is 0.013 times that of the earth. 
Taking the earth's radius as 4000 miles and the distance of the 
moon's center from the earth's center as 240,000 miles, find the 
distance of the center of gravity of the earth and moon together 
from the center of the earth. 

179. From a right circular cylinder of height h there is cut 
a cone of revolution whose base coincides with that of the cylin- 
der, so that the center of gravity of the remaining solid coincides 
with the vertex of the cone. Find the altitude of the cone. 

180. Find the distance above the lower base of the center of 
gravity of a trapezoid having bases of length a and b, and alti- 
tude li. 



Cextek of Gravity 



111 



181. Find the center of gravity of a rifle bullet consisting of a 
cylinder 2 calibers in length and a paraboloid 3/2 calibers in 
length,, having a common base, the opposite end of the cylinder 
containing a conical cavity 1 caliber in depth, with base equal in 
size to that of the cylinder. ( Caliber = diameter of shell.) 

182. A casting is in the form of a hollow cylinder with one end 
closed. The thickness of the end is 1 inch, the length of the cast- 
ing 12 inches, and the radii of the inner and onter surfaces of 
the cylinder are 5 inches and 6 inches. Find the position of the 
center of gravity of the casting. 

183. If the casting described in the preceding problem is 
filled with a material one-fonrth as heavy as the material of the 
casting, find the position of the center of gravity. 

184. A solid consists of a hemisphere of radins a and a cone 
of revolution, having a common base. The altitude of the cone 
is a. Find the distance of the center of gravity from the common 
base. 

185. Derive the formulie for Xa and Xo for the centers of grav- 
ity of the plane sections of beams given on pages 8-1-86 of Hud- 
son's Engineer s Manual. 

62. The Theorems of Pappus and Guldinus. — First Theorem. 
— The volume of any solid generated by the revolution of a plane 
area about an external axis is equal to the product of the area o'l 
the generating figure and the distance its center of gravitv 
moves. 




Fig. 129. 



Second Theorem. — The area of any surface generated by the 
revolution of a plane curve about an external axis is equal to the 



112 



Elementary Mechanics 



product of the length of the generating curve and the distance 
its center of gravity moves. 

To prove the first theorem: Let the figure MN (see Fig. 129), 
with area A and center of gravity at {x, y), revolve through an 
angle a about OX, which does not intersect the area MN. Let 
dA, at the point (x, y), denote an element of area in MN , and 
dV the corresponding element of volume, obtained by rotating 
dA about OX through the distance ya. Then 

dV = ya'dA, 
whence 



Y 



But 



= a ^ dA, 

MN 



ydA = y^A, 



MN 



therefore, 

which was to be proved. 



V = ya'A, 




Fig. 130. 



To prove the second theorem: Let the plane curve MN (see 
Fig. 130), of length I and center of gravity at (x, y) , revolve 
through an angle a about OX, which does not intersect MN 



Center of Gravity 113 

except possibly at its end points. Let ds, at the point {x, y) , 
denote an element of the arc MN, and dS the corresponding 
element of surface generated by rotating ds about OX through 
the distance ya. Then 

dS = ya ds, 
whence 



But 



8 = 

MN 



yds = y'l. 



ay ds. 



therefore, 

S = ya-l, 

which was to be proved. 

Note that both theorems are valid when a has any value from 
to Stt. 

Problems 

186. An equilateral triangle revolves about its base, whose 
length is a. Find the surface and the volume generated. 

187. Find the surface and the volume of the anchor ring 
generated by a circle of radius a* making a complete revolution 
about an axis 6 units distant from the center of the circle (&>a). 

188. Using the known values for the surface and the volume 
of a sphere, find the position of the center of gravity of the arc, 
and of the area, of a semicircle. 

189. Find the volume obtained by revolving about its longest 
side a triangle whose sides are 6, 8, and 10. 

190. Using the results of Ex. 5, Art. 60, find the volume of 
an ellipsoid of revolution. 

191. Using the result of Prob. 168, find the volume generated 
by revolving about its asymptote the area between the cissoid and 
its asymptote. 

63. Approximate Integration. — Hitherto we have solved only 
problems in which it was possible to express the function y as 



114 Elementary Mechanics 

an explicit formula in terms of x, and in which, moreover, it 
was always possible to find the indefinite integral of y. But 
many cases arise in practice for which one or the other of these 
conditions is not satisfied. An example of the first sort is to 
find the water-line area of a ship from a careful drawing when 
the equation of the bounding curve is not known and cannot be 
written in any simple form; an example of the second sort of 
difficulty arises in finding the quantity of water flowing per 
second through a circular outlet of radius r, whose center is h 
feet below the surface of the water. A solution of this last prob- 
lem involves the evaluation of 



y{r'-x^){h-x) dx, 

and this, called an elliptic integral, cannot be expressed in terms 
of the functions treated in the elementary Calculus. In such 
cases we must resort to approximate methods for the evaluation 
of the definite integrcd. 

To derive an approximate formula, graph y as a function of x, 
obtaining BoBn (see Fig. 131). Then restricting y not to change 
simi between x = a and .^ = 5, we have 



'to 

r 



y dT = area AoBoBnAn. 



Divide this interval from x — a to x — l) into n equal parts li — 
{!) — a) / n in length, and at the points of division Aq,A^, . . . . , 
^n. erect ordinates ?/o, 2/i^ - - - - , ^h, meeting the curve in i?o. -Bi, 
. . . . , Bn, respectively. The points of division must be so 
chosen that the values of the ordinates can be obtained from 
available data, by measurement or by computation. An obvious 
first approximation to the area AoBn is obtained by replacing the 
arcs BqB^, B-^B^, . . . . , Bn.^Bn, by their chords and then add- 
ing the areas of the trapezoids AqBoB-^A^, etc. This is some- 



Center of Gravity 



ii; 



times used, under the name of the " Trapezoidal Eule." A better 
approximation can usually be obtained by taking the points B in 
sets of three or more and passing through each set of points a curve 
which, for the small distance usually taken, very nearly coincides 
with the corresponding piece of the given curve, and the area 
under which can be found by the usual methods of the Calculus. 



B. B, 



(a.o) 



B... 



B„.. 



B„ 



3n-. 



(b,o) 



Aq a, 



A. A. 



Fig. 131. 



64. Simpson's Eule. — One of the most useful of these approxi- 
mate formulas, known as " Simpson's Eule,'' requires that the 
number, n, of divisions of AoAn shall be even, and approximates 
the curve through each set of three successive points B by an arc 
of a cubic curve of the form 



y — a+'bx + cx^ + dx^. 



116 



Elementary Mechanics 




But 



To find the area between each segment of the arc and the X-axis, 
suppose the F-axis to be translated to coincide with the second 
of the three successive ordinates yo, y^, 2/2^ spaced li units apart. 
(See Fig. 132.) 

We have 

Area ^0^2= ydx=\ (a + hx-^- 

J-h J-h 

cx^-\-dx^) dx 
= {h/3){6a + 2ch^). 

y^ — a — hh-^- ch^ — dh^, ' 

^i = «. 

y^ = a+bh + ch^ -T dL^, 
A2 whence 

yo + 4?/i + 2/2 = 6a. + ^ch"^. 
Therefore, 

area strip AoBoB^A^^^lifd) {yo + ^^ + y2)' 
area strip A^BJB^A^—{h/2)){yo + 4:y^ + y4). 

area strip ^„_25n_25„^n= (/i/3) {yni^ + ^n-^ + yn). 
Adding, we obtain, as an approximate value for y dx, 

(V3)(2/o + 4?/, + 2?/. + 47/3+ .... +22/»-2 + 42/n_i + ?/n). 

This is Simpson's formula. While we have used the area 
under a curve in obtaining this result it should be understood 
clearly that it is not merely a device for approximating such 
areas, but that it is rather a method for approximating the value 

of the definite integral y dx. 



Center of Gravity 117 

65. Illustrative Example 

Find the height of the center of gravity above the lower base of 
the solid shown in Fig. 133. The solid is 24 feet high, and the 
five parallel sections whose areas in square feet are given, are 
6 feet apart. \Ye have to compute 




Fig. 1 



For the numerator 2j = x A, for the denominator y = A, and 
]i = 6. The compntation can profitably be arranged in tabular 
form, as follows: 


A 

300 
260 
206 
142 

74 


c 
1 
4 
2 
4 
1 


cA x/h 
300 
1040 1 
412 2 
568 3 
74 4 


cAx/h 



1040 

824 

1704 

296 




12 


2394 


3864 


5; = 


. sum of cA: 


_ (h/o) (sum of 
(h/3) (sum of 


cAx) 
cA) 




r_ 3864x6 _q 


.68 feet. 



sumofc.4 2394 

Instead of using values 'of x as multipliers in the fourth column, 
to lessen the computation we have used the values of x/h and 
afterward multiplied the sum of the last column by h. 



118 Elementary Mechanics 

Note that the column c reads the same up or down and its sum 
is 3n. If we count the number of times each ordinate is repeated, 
the sum of cA is the sum of 3n ordinates. Referring to Fig. 131, 
the base equals hn, and 

CD 

area= y dx= (h/ 3) (sum oi cy) 

= (/m) X (sum of Sn ordinates) /3n 
= base (average ordinate), 
or again, 

= base (sum of cy)/{sum of c). 

Thus in the example the volume of the solid is given by 



i: 



A tZa; = 24.2394/12=:4788 cubic feet. 

Problems 

192. Find the area bounded by the curve 8y = x^, the X-axis 
and the lines x = 2 and a; = 4. Check the result by integration. 
Under what conditions on y will Simpson's Eule, for n = 2, give 

the exact value of y dx? 

193. Find the area between a segment of a curve and the X- 
axis if ordinates taken 4 feet apart have lengths 3.6, 4.2, 4.7, 5.1, 
5.4, 5.6, 5.7 (all in feet). Find the distance of the center of 
gravity from the left ordinate. 

194. Find the length of the perimeter of the ellipse a: = 4cos</), 
?/ = 3 sin <^, taking /i = 1 5 ° . 

195. The quantity of water flowing per second through a cir- 
cular orifice of radius r whose center is constantly h feet below 



the surface of the water is 9.76 1 V {r'^ — x~) {li — x)dx, r and h 

being given in feet. Graph the integrand as a function of x, 
and compute the integral, when r = l foot and /i = 2 feet. 

196. A solid 20 feet high has a rectangular base 20 x 22, sec- 
tions parallel to the base at intervals of 5 feet are rectangles 
15 X 17, 11 X 13, 8 X 10, and 6 X 8 (all in feet). Find the volume 
of the solid, and the distance of the center of gravity above the 
base. 

197. Find the area and the center of gravity of the cross 
section of a steel rail, 6 inches high, symmetrical about a vertical 



CkXTKK of tiKAVlTY 



119 



axis aiul whoso width in iiu'hos taken povpenclieuhir to this axis 
at halt'-iuch intervals, bei^innino- at the base^ is rcspeetively G, 
5.5, 3.23, 1.24, 1.0, l.O, 1.0, 1.18, 1.24, 2.5, 4.24, 4.08, 4. 

198. The water-line area of a ship is 300 feet long, and half- 
ehords of this area, measured perpendicular to its longitudinal 
axis at intervals of 12.5 feet, are found to be 0, 12, 18.5, 22.5, 
25.5, 27, 29, 29.5, 30, 30.5, 31, 31.5, 32, 31.5, 31, 30.5, 29.5, 
28.5, 27, 24, 19.5, 14, 8..\ 3, 0.5. Find the water-line area. 




B 



Fic. 134. 



\99. Cross soetions o\^ a ship, taken at intervals of '24 feet, 
have iho areas in square feet li^O, o60, 480, 553.9, 400, 140, and 
0. Find the volume and the distanee of the center of gravity 
froni the stern. 

'200. Find the area of the section .4^. the volume of the bunker, 
and the distance of the center of gravity from the end section 
AJl, from the data given in Fig. 134. The sections A^, A.., A^, 
A^, A^ are perpendicular to A^Ar,, and the distances in A^^B 
are measured perpendicularly to A-^^B^. 



120 



Elementary Mechanics 



^2 = 4220 square feet. 

^3 = 3740 square feet. 

A^ = 3020 square feet. 

^5 = 2020 square feet. 



Moment of Inertia. 



-The integral. 



r-dm, where r is the 



distance of an element of mass dm from an axis, is of such fre- 
quent occurrence in mechanics and its applications that a pre- 
liminary study of its computation is desirable. We shall show 
later (Art. 172) that the moment which measures the resistance 
that a body offers, through its inertia, to change in angular 
velocity, is proportional to this quantity ; hence the name 

Moment of Inertia = /= \r~dm. 

M 

It follows at once from the definition that the moment of inertia 
of a composite body is equal to the sum of the moments of inertia 
of the pieces into which the body can be broken up. In particyi- 
lar, if the body is made up of a number of particles, the integral 
sign becomes a simple summation. 

67. Illustrative Examples 

Example 1. — Find the moment of inertia about one side, of a 
thin rectangular plate of constant density. 

Let the plate have dimensions 5 by /i.- Y C 

Take two intersecting sides of the plate 
as axes, as indicated in Fig. 135, and find 
la;. Take as the element of area a slice dy 
thick, parallel to the a;-axis. Then 

dm = pdA — pJ) dy. 



r~dm — 



if pi dy — pl)U^/o. 



But 

M^phh, 
therefore 

7^ = ilf /i73. 



dA 



Or 



da 



Fig. 135. 



Cexter of Gravity 



121 



Xote that moment of inertia is measured in a hitherto imfa- 
mihar unit having dimensions Mass- (length)^; for example in 
pound-inch- units, ton-foot- units, etc. Thus if the rectangle of 
Example 1 has a mass of 20 pounds and ]i = 6 inches, its moment 
of inertia I^ = M ■]i-/S = 20 -6-/3 = 24:0 pound-inch- units. In 
many applications we are concerned with another quantit}' 
^}'-dA. The dimensions of this are clearly (length)*, length 
entering to the second degree on account of 7*^ and again on 
account of ^.1. This quantity has nothing to do with inertia, 
but is nevertheless commonly called a " moment of inertia " and 
denoted by I, on account of its similarity in form to the true 
moment of inertia. This quantity taken for any plane area is 
the same as the true moment of inertia of a flat plate of unit 
density and unit thickness occupying the same area, for in this 
case we should have dM=dA. Thus the moment of inertia of 
the area of a rectangle 6 inches by 8 inches about the 8-inch side 
is 8x6x6-/3 = 576" inch* units. 

Example 2. — Find the moment of inertia of a homogeneous 
thin circular plate about an axis per- 
pendicular to the plane of the circle 
through its center. 

Take a ring-shaped element of area 
of radius r, as indicated in Fig. 136. 
Denote the radius of the oiven circle 
bvfl. Then 



and 



dm — p27rr dr, 



/=[ r-dm=2iTp^ rHr 
= 7rpa^/2=:\Ia-/2. 




Fig. 136. 



Similarly the moment of inertia about its geometric axis of a 
homogeneous right circular cylinder having this circle as a base 
is also If a-/2. 

Example 3. — Find the moment of inertia of any homogeneous 
solid of revolution about its axis. 

Take as X-axis the axis of symmetry for the sohd, cutting its 
bases in the points {a, 0, 0) and (&, 6, 0). Take as the infini- 
tesimal element of mass a cylindrical slice with radius y and 



122 



Elementary Mechanics 



thickness dx (Fig. 137), whose moment of inertia about OX, 
as found in the preceding example, is 




whence. 



Fig. 137. 
dh = p{7ry-dx)(y~/2), 



I.= 



fb''- 



Example ^. — Find, using Simpson's Eule, the moment of 
inertia about OB of the area OBCA 
(Fig. 138), all dimensions being in 
feet. Here 




dA = y dx, p = l, 



whence 

Ion 



=z r- dm = X' y dx. 

JM Jo -^ 



For Simpson's Rule, /m=4, w = 4, and 
x's have a common factor 4. Hence 
we have 

(4/3) 726x16 = 15,488 (feet)*. 



Fig. 138. 












i;V16 


y 




x'y/lQ 


c 


cxHj/lQ 





20 







1 





1 


18 




18 


4 


72 


4 


15 




60 


2 


120 


9 


11. 


5 


103.5 


4 


414 


16 


7. 


5 


120 


1 


120 
726 



Center of Gravity 



12: 



right 



circular cone 



ProMems 

201. Find the moment of inertia about its base of a triangle 
of altitude h and base b. 

202. Find the moment of inertia of a 
about its geometric axis. 

203. Find the moment of inertia about a diameter of a homo- 
geneous sphere of radius a. 

204. Find the moment of inertia of the area bounded by the 
cardioid r = fl(l + cos </>), (a) about the initial line as an axis, 
(&) about a line perpendicular to the plane of the area, through 
the point (0, tt). 

205. Find h for the area bounded by the astroid x-/^-\-if/^ = 

206. Find the moment of inertia about its major axis of the 
area bounded by an ellipse with semi-axes a and 5. 

207. Find the moment of inertia of the area bounded by the 
parabola y^—{'b^/a)x and the line x^a, about the axis of the 
area; also about the tangent to the parabola at the vertex. 

208. Find the moment of inertia about the X-axis of the solid 
obtained by revolving the area of the preceding problem about 
the Z-axis. 

68. Theorem of Parallel Axes. — AVhen the moment of inertia 
of a body about any axis is known, its moment of inertia about 
any parallel axis can be obtained without further integration by 
using the following important theorem : 

If the moment of inertia about 
any axis I, of a body whose mass 
is M be denoted by Ii, and its 
moment of inertia about a parallel 
axis through its center of gravity 
be denoted by Iq, then 



h = Io + dUI, 



where d is 
the axes. 

Pass a plane 



the distance beliveen 



through G per- 




dm 



^0 



,'<^ 



Fig. 139. 



pendicular to the parallel axes. 



124 Elementary Mechanics 

cutting / in (Fig. 139). Take GO as the A'-axis in a system of 
coordinates, with origin at G. Let A be the projection on this 
plane of any element of mass dm. Let GO — d, GA=r, A0 — 7\, 
OGA — <^. Then, using the Law of Cosines, from Plane Trigo- 
nometry, 

r - dm = (7'~-^d^ — 2rd cos d> ) dm 

M J .V 

■r dm + d'- dm — 2d\ r cos W) din. 

M J .V JM 

But /■ cos cji — x, whence 



Ii 



r cos (/) dm 

M 



M 



X dm = xM, and x = 0, 



since G is the center of gravity for M. Further, 
r- dm = Iq, dm = ilf . 

]}1 J M 

Therefore 

Corollary: If I and ?i are two parallel lines distant respectively 
d and c?i from a parallel axis through the center of gravity of a 
body of mass M, then 

Ii-dm=^lG = Ii^-d^^M. 

69. Radius of Gyration. — From its definition, moment of 
inertia must have the dimensions (length)- X mass, and hence 
can be expressed in the form 

This defines a length Tc, measured from the axis of rotation, and 
called the radius of gyration. Just as we have previously replaced 
the turning moment about any axis of the gravitational forces 
acting on a body with mass M, by the moment due to a particle 
endowed with the same mass and located at the center of gravity 
of the body; so here we replace the moment of inertia I of any 



Center of Gravity 



125 



mass by that of a particle or system of particles having together 
the same mass and distributed in any convenient manner on the 
surface of a circular cylinder whose radius is Ic and whose axis 
is the axis of rotation of the given mass. 

Substituting I = h~M for / in the theorem of parallel axes, 
and cancelling M, we obtain 

Jcr^^Ji-G^ + d^. 

70. lUusiraiive Examples 

Example 1. — Find the radius of gyration for Example 1, and 
for Example 2 of Art. 67. 

In Ex. 1, h = MJr/3, therefore ^,- = 7A¥ = 7i73. 
In Ex. 2, I = M a-/2, therefore k- = I/M = a~/2. 




Fig. 140. 



Example 2. — Find /t-- for a triangle of base h and altitude h 
about the base as an axis; also about a line through the center 
of gravity of the triangle, parallel to the base. 

In solving Prob. 201, it was found that the moment of inertia 
of the triangle about its base is 5 •/i^/12. M = area of triangle = 
1)11/2; therefore Icj;':=zl^/M^li^/Q. The distance d between the 
base and a parallel axis through the center of gravity is d — h/o 
(Fig. 140). Therefore 

A-g' = hr - ^- = h-/6 - /i79 = /i V36. 

Example 3. — Find the moment of inertia of a rectangle about 
a gravity axis parallel to one side. 



126 



Elementary Mechanics 



2"l 



Denote by /& and Iq respectively the moments of inertia about 
the base of the rectangle, and about a parallel axis through the 
center of gravity, tliese axes being d = h/2 apart. We showed in 
Example 1 that I'^^ — lr/S. Therefore 

]co~ = h' -d- = lr/3 - ]i-/4: = ^712, 
and 

I = h^-M={V-/12)hh=zhhyi2. 

Example U- — Find the moment of inertia of the plane section 
of a beam shown in Fig. 141 about an axis 
through its center of gravity, parallel to 
the shorter side. 

The center of gravity is computed to 
be 5 inches above the base of the section. 
Divide the area into the rectangles Ijc, 
Id, and de. Then, about A A we have for 
areas he, hd, de, respectively, A;- = 49/3, 
25/3, 16 + 4/1^- 

The moment of inertia of the entire A 
section is equal to the sum of the moments 
of inertia of the three pieces, that is, to 
the sum of the products of each h^ by 
the corresponding area. Hence, 

1= (49/3)14+ (25/3)10 

- + (49/3)6 = 410 inch^ 



G|,. 



lO' 



I d 5 
Fig. 141. 



A 








3 


G 


3" 
10 


3 


^2-> 


3" 











8 

Fig. 142. 



Example 5. — Find the moment of 
inertia about AA of the plane section 
of an I-beam shown in Fis^. 142. 

We may think of this area as that 
of a rectangle 8'' by 12" from which 
two pieces, each 3" by 10", have been 
cut. Hence, using in each case the 
moment of inertia of a rectangle about 
a gravity axis parallel to one side, we 
have 

Iaa =8 12712-2(3 10712) 

= 1152-500 = 652 in.^ 
/tUA' = 652/36 = 18.11 in.2. 



Center of Gravity 



127 



Problems 

Find I and Ic^ for each of the following bodies (Problems 209 
to 220) about the axis indicated: 

209. The Tee-section shown in Fig. 143 about AB. 



15 



18 



B 



10 



Fig. 143. 



Fig. 144. 



210. The angle section shown in Fig. 144 about an axis 
through the center of gravity, parallel to AB. 

211. A hollow rectangle, outside dimensions h by h, inside 
dimensions b-^ by h-^, about a gravity axis parallel to the base. 

212. A hollow circle, outer radius R, inner radius r^ about a 
diameter. 

213. A hollow right circular 
cylinder, outer base radius K, 
inner base radius r, about its 
geometric axis. 

214. The one base frustum of a 
paraboloid of revolution about its 
axis. 

215. A uniform rod of length 
h about an axis perpendicular to 
the rod at one end. 

216. A semi-circumference 
about its diameter. 

217. A semi-circular area about 
a tangent at one end of its -diam- 
eter. 

218. The plane sections of 




Fig. 145. 



128 



Elementary Mechanics 



Hudson's Manual, pages 84, 85, 86, about the axes AA and BB. 

219. The cross section of the steel rail shown in Example 5, 
Art. 70, about its base. 

220. The area OABC about OA, where 0C = 25 feet, and seven 
equally spaced ordinates, beginning with OA and ending with CB, 
are found by measurement to have the values 27.5, 26.5, 25, 23, 
19.5, 14.5, and 5, all m feet (Fig. 145). 

71. Polar Moment of Inertia. — The moment of inertia of a 

thin plate (or plane area) about an 
axis perpendicular to its plane is 
called its polar moment of inertia, 
and is denoted by Ip. Take the 
given axis, intersecting the given 
plane in (Fig. 146), as the Z-axis 
in a system of rectangular coordi- 
nates, and any two perpendicular 
lines through 0, in the given plane, 
as X~ and Z-axes. Then 




Fig. 146. 
r^ dm = 



Therefore 






Ir 



Since OX and OY are any two perpendicular lines through 
in the plane, it follows that the sum of the moments of inertia of 
a plane area, with respect to two rectangular axes in its plane, is 
the same as the sum of the moments of inertia ot tins area with 
respect to any other two rectangular axes in the same plane 
passing through the same point. Consequently, if Ix is the maxi- 
mum value of the moments of inertia about all lines through 
in the plane, ly must be the minimum value (and vice versa), 
since their sum is constant. Two axes OX and OY having this 
property, that Ix is a maximum and ly a minimum, are called 
principal axes for the point 0, and Ix and ly are called principal 
moments of inertia for 0. 



Center of Gravity 129 

It can be shown that an axis of symmetry for the area, when it 
exists, is a principal axis. Further, if we take this axis of sym- 
metry for OX, and for OY the perpendicular axis through the 
center of gravity of the area, then the smaller of Ix and ly will 
be the least moment of inertia that can be obtained for the area 
about any axis in its plane. 

72. Illustrative Examples 

Example 1. — Find the moment of inertia of the area of a 
circle about a diameter, and about a polar axis through a point 
on the circumference. 

It was shown in Ex. 2 of Art. 67, that the polar moment of a 
circle about an axis through its center is Ma'^/2. Hence the 
polar moment about a parallel axis through a point on the cir- 
cumference' (a units away) is 

Ip^ = 21 a- /2 + Ma~ = 3Ma-/2. 

Because of the symmetry of the circle, its moment of inertia Id 
about a diameter is the same for all diameters. Hence if we take 
a set of axes in the plane of the circle, with origin at its center, 
then 

Ij:=Iij = ID' 

But 

Ip=Ix + Ii/ = 2 Id J 
and 

Ip=]\Ia'/2. 
Therefore 

Example 2. — Find the moment of inertia of a homogeneous 
solid of revolution about an axis perpendicular to its axis of 
rotation. 

Take the axis of rotation for the solid (Fig. 147) as the X- 
axis, and find ly. Take as the element of mass dm a slice dx 
thick perpendicular to the Z-axis. Then dm^piry^dx. Let AB, 
parallel to OY, be a diameter of the element of volume. Then, 
from the preceding example. 



130 



Elementary Mechanics 



d{lAB) = {p7r2fdx)if/4:. 

From the theorem of parallel axes, 

d{Iy) =d{lAB) +x^dm = p7ry^{y~/4:-^x^)dx. 

ly is obtained by integrating this between the proper limits. 

Example 3. — Find the moment of inertia about its axis of a 
right circular cylinder four feet in diameter, weighing 160 
pounds. 

Problems of this type illustrate the importance of expressing 
the moment of inertia of homogeneous solids in the form I = Mk^. 



dV. 



'; 


5 


H 

Iclx 1 


(b.«o 




/ A' 


\J 




Fig. 


1^ 


17. 


\j 





Thus, for the cylinder in question, ]v^=:a^/2 = 2, and I — 2M, 
where the value of M will depend on the units of mass being 
used in a discussion in which this problem might arise. For use 
in most problems in the dynamics of a rigid body it will be found 
convenient to use (see Art. 122) the "engineer's unit of mass'* 
or " g-pound," which requires that a body whose weight is W 
pounds have a mass M=W/g, where ^ is a constant for all 
bodies at the same place on the earth's surface. For the purposes 
of the present chapter we shall take (7 = 32; it will be seen later 
that this requires that time shall be expressed in seconds, and 
length in feet in any equation in which this determination of 
mass enters. Thus for the present problem, ilf = 160/32 = 5 
'' g-pounds," and 7 = 5x2 = 10 g~pound f eet^ 



Center of Gravity 131 

Pro'blems 

Find I and Tc"^ for each of the following bodies about the axis 
indicated. 

221. A cube about a gravity axis parallel to an edge. 

222. A rectangular prism about a gravity axis parallel to an 
edge. 

223. An anchor ring obtained by rotating a circle of radius r 
so that its center traces the circumference of a circle of radius B; 
about the axis of rotation ; about a diameter perpendicular to this. 

224. A right circular cone with altitude li and base radius a, 
about an axis through the vertex parallel to the base; about a 
gravity axis parallel to the base. 

225. The area of the lemniscate r- = a-cos2(/> about a polar 
axis through its center. 

226. The area between the parabola if = a{a — x) and the 
circle x^ + y^ = a^ about a polar axis through the origin. 

227. The head of a projectile is formed by the revolution of a 
semi-parabola about the ordinate h, so that the height h is the 
radius of the base, or one-half the caliber a; about the axis of 
rotation. 

228. A body consisting of a hemisphere and a cone of revolu- 
tion of the same base, and height equal to the radius; about an 
axis through the vertex, parallel to the common base. 

229. The ellipsoid xya^-i-y^/V-\- 2^6^ = 1, about the Z-axis. 

230. A circle of radius a, the density varying as the square of 
the distance from the center ; about a diameter. 

231. The plane section of a beam is a rectangle 6 inches by 10 
inches. Find its least moment of inertia. 

232. The plane section of a column is an ellipse with semi- 
axes 6 inches and 8 inches. Find its least moment of inertia. 

233. Find the moment of inertia about a diameter of a homo- 
geneous sphere 18 inches in diameter, and weighing 96 pounds. 

234. Find the moment of inertia about its axis of a hollow 
cylindrical drum, weighing 120 pounds, and having an inner 
radius of 18 inches, an outer radius of 24 inches. 



132 Elementary Mechanics 

CHAPTER YI 
FLUID PRESSURE 

73. Properties of Fluids. — Up to this point we have been con- 
cerned only with rigid bodies, i. e., bodies which do not change 
their shape or volume when acted upon by external forces. This 
is largely an assumption, since in reality, if sufficient force is 
exerted, the shape or volume, or both, of any body will change. 
However, for practical purposes, it has been necessary and suf- 
ficiently accurate for us to assume that these bodies were rigid. 

Matter exists in the form of either a solid, a liquid, or a gas, 
although sufficient change in temperature and pressure will 
cause any substance to pass from one of these forms to another. 
A solid has a definite form and volume and offers marked resist- 
ance to any change of these; a liquid has a definite volume but 
takes the form of the vessel containing it ; a gas has neither defi- 
nite form nor volume but expands till it fills the vessel contain- 
ing it. 

We define a fluid as a body which offers no permanent resistance 
to forces which tend to change its shape. 

A " perfect fluid " is frictionless, and can exert no resistance 
to shear or tensional stress. (See chapter on " Def omiable 
Bodies," Arts. 84 and 89.) Such a fluid takes the exact form of 
a containing vessel. Actually, fluids do offer a slight resistance 
to distortion of form, and the measure of this resistance is the 
" viscosity " of the fluid. 

Water at rest, which is the case in Hydrostatics, may be con- 
sidered, with negligible error, as a perfect fluid. 

Fluids can, broadly speaking, be divided into two general 
classes, liquids and gases. 

A liquid is a fluid which changes volume very slightly under 
compression. 



Fluid Pressure 133 

A gas is a fluid whose volume changes sensibly under com- 
pression. 

All fluids have weight. In the case of liquids in an open vessel 
the thing of importance is the force due to the weight of the liquid, 
and its effect upon the walls of the container or upon the surfaces 
of submerged bodies. On the other hand, a gas confined in a closed 
vessel completely fills the vessel, and the forces acting upon the 
walls of the vessel are due primarily to the pressure of the gas 
under compression, and not to the weight of the gas, the force 
produced by the latter being relatively negligible. 

74. Pressure. — Theorem. — " When a fluid is at rest, the force 
which it exerts on any surface with which it is in contact, is nor- 
mal to that surface." 

For, if the force were not entirely normal it would have an 
unbalanced component tangent to the surface, and, since the 
fluid cannot resist this tangential force, motion would take place, 
which is contrary to hypothesis. Thus, in the Fig. 148 is rep- 



IB^— 



Fig. 148. 

resented a plane AB submerged in a liquid. If the force F at 
the point P were not normal, i. e., not perpendicular to the plane, 
there would be an unbalanced component of F in the direction 
parallel to the plane which could not be resisted by the fluid. 

Liquids in an open vessel present a " free surface " to the air. 
An immediate consequence of the above theorem is " that the free 
surface of a liquid at rest is perpendicular to the force of gravity, 
and so is practically a plane." 



134 Elementary Mechanics 

In considering the forces exerted by liquids with a free surface, 
we shall neglect the force due to the weight of the air on this sur- 
face, unless otherwise specified. 

If a plane surface be submerged horizontally in a liquid, to a 
depth of h feet below the free surface, then on every square foot 
of this plane there is a force exerted normal to it, of 

(hw) pounds, 

where w is the weight per cubic foot of the liquid. The distance 
h is called the " head " on this plane. This result rests on the 
assumption that the density of the liquid is constant. For the 
small depths usually encountered in hydrostatic problems, this 
assumption produces only a negligible error. If the area of the 
horizontal plane is A square feet, then the total force F, exerted 
by the liquid on the plane, is 

F — Aliw pounds. 

The intensity of the force is uniform over the horizontal plane 
surface, i. e., is the same at every point. If F is the total force, 
and A the area, then 

p = F/A 

is defined as the " pressure " on the horizontal plane, and is given 
in pounds per square foot, or some similar units. In the case of 
the horizontal plane, the pressure is the same at all points of the 
plajie. 

75. Theorem. — " The pressure (force per unit area) at any 
point of a fluid at rest, is the same in all 
directions." 

Consider a portion of the fluid, in the 
form of a right triangular prism (Fig. 
149), the legs of the right angle being 
horizontal and vertical as in the figure. 
Let F be the force on the horizontal face. 
Then 

j~-T- = p. or F=:p-dx.dz, 
Fig. 149. "^ * "^ 

where p is the pressure on that face. As 
dx and dz approach zero, p remains constant. This force F 
must be balanced by the vertical component of N, the force 




Fluid Pressure 135 

normal to the inclined face, since on all the other faces the forces 
have no vertical components. This vertical component is N cos 6^ 
and hence 

F=N ' cos 0. 

Let p' be the pressure on the inclined face at its upper edge. 
This pressure increases as we go down the inclined face, due to 
the increasing heads on this surface. If e is the increase in pres- 
sure at the lower edge, then 

p' 'dZ'ds<N<(p' + €) -dz-ds. 
But 

ds • cos = dx, 
and so 

p' 'dx-dz<CF<C (l?' + e) 'dx-dz 

.*. p'<p<p'-\-e. 

Now as the prism shrinks to a point P by letting dx, dy, and dz 
approach zero, then € = 0, and so, at the limit, 

p' = p. 

Eesolving horizontally, parallel to dx, we find the limit of the 
pressure on the face dy dz, the same as that on the inclined face, 
and so is equal to p. Since 6 may be any angle, this proves that 
the pressure at any point is the same in all directions. 

76. If we now consider a plane of finite extent, submerged and 
inclined to the horizontal, the pressure at a point P, at depth li, 
is then the same as if the plane were turned to a horizontal posi- 
tion at depth h, and so 

V = F/A, 

where here F is the force exerted on any area A at constant 
depth h. In the case of the inclined plane, p varies from point 
to point on the plane, but satisfies the relation 

p — iuli. 

In general, if any surface be submerged, the pressure at any 
point at depth li is given by 



136 Elementaey Mechanics 

and the force producing this pressure is normal to the surface at 
each point. 

77. The Transmissihility of Pressure. — Suppose a fluid com- 
pletely fills a closed vessel, and a constant force be applied to a 
piston in the walls of the vessel (Fig. 150). If the force tends 
to push the piston in, the fluid becomes compressed. Since the 
fluid and the piston are at rest, the force on the outside produces 
a definite pressure on the fluid immediately adjacent to the piston 
head, which pressure must be transmitted without change to every 
l^oint on the inside of the walls of the vessel. To prove this 
statement, assume that the force on the piston is the only force 
acting on the fluid, and consider a right circular cylinder of the 
fluid extending normally from a unit area at A on the piston to 
a point P in the fluid. The forces acting on the curved surface 
of this cylinder are normal to that surface and so have no com- 




FiG. 150. 

ponent along the axis of the cylinder. Since the cylinder is at 
rest, then the forces acting normal to its ends are in equlibrium, 
and so the pressure at P is the same as that at A. Now, since 
the pressure at P is the same in all directions, we may turn the 
cylinder about P so as to end at or near any point B of the walls 
of the vessel, and so conclude that the normal pressure at B is 
the same as that at P and so is the same as that at A. 

This pressure is of course in excess of that due to the weight 
of the fluid, the latter varying with the depth. 



Fluid Peessure 



137 



The principle involved in the above discussion is called that of 
" the transmissibility of pressure within a fluid." The hydro- 
static press is an application of this principle. 

Example. — The lever of a h3'drostatic press is 3 feet long 
(Fig, 151). The distance from its fulcrum to where it acts on a 




Fig. 151. 



small piston is 8 inches. The diameters of the pistons are -| inch 
and 1 foot. If 1 pound is applied at the end of the lever, what 
weight (W) will the large piston just lift? 

Here the eliective force at the small piston is found, by taking 
moments, to be 

36/8 pounds. 

Hence the pressure produced in the water is 

.•. 17 = 2593 lbs. 
78. Total Pressure Force. — We have seen that the pressure at 
a point of a submerged surface is given by 

p = iv]i. 

If the surface is horizontal, then the pressure is constant, and 
the total force, F, is given by 

F = iuliA lbs. 
In the general case the surface is not horizontal, and the pres- 
sure varies from point to point. If the surface is plane, the force 



138 



Elementary Mechanics 




vectors at each point are parallel, and so the total force has a 
physical meaning. It is the thrust on the plane clue to the liquid 
pressure, or the total pressure force, and it is normal to the plane. 
Its point of application will be considered later. 

Consider any plane area A (Fig 
152), submerged at an angle a with 
the horizontal, and let I be the line of 
intersection of the plane of A and the 
free surface. Take an element of area 
AA, in the form of a strip whose edges 
are parallel to the surface of the liquid, 
.AA and at distances 

y and y-\-Ay 

from I. Then the element of force on 
Fig. 152. ^-^ is w • ^ sin a • AA pounds ; so the 

total force on A is given by 

F = lim {Ay = 0) '^Aif^ -y sin a- AA=\ w-ysma-dA, 

where the integration covers the area A. But we observe here that 

F=\ w -sin a- ydA — w -sin a\ y • dA = w -A- {y -sma), 

where y • sin a is the distance of the center of area A below the 
surface of the liquid, or y is the distance from l, in the plane A, 
to the center of area. Hence we have the 

Theorem. — " The total pressure force on a submerged plane 
area is equal to the product of the area, the weight per unit 
volume of the liquid, and the depth of the center of area." 

Examples 

1. Find the thrust on a vertical circle of radius 2 feet, sub- 
merged in water with its center 10 feet below the surface. 

Here iu = 62.5 Ibs./cu. ft. A — A:tt sq. ft. and ?/ • sin a =10 ft. 
.*. i^= (62.5) . (47r) . (10) =7860 lbs. 



Fluid Pressuee 



139 



2. Compute the total pressure force on a vertically submerged 
parabolic segment, of width 2h feet and depth b feet, with upper 
edge in the surface of the liquid (Fig. 153). 




\\hy^ 



.tf^)dy = j\wlih~ lbs. 



h-ij).2x.dy. 

J o 

But here 

x^ : h'^ = y: b : 
2wh 

Vb 

3. A rectangle hxb feet is submerged with its edge 7i in the 
surface, and its plane turned 30° from the vertical. Determine 
the total pressure force on this rectangle. 

Here A = bh, a=:60°, and y = ^. Hence F=lwb-h\/3 pounds. 

79. Cylindrical Surfaces. — We will now consider the total 
pressure force in a given direction 
on C3'lindrical surfaces. A problem 
of practical importance is to deter- 
mine the stresses set up in the mate- 
rial of a steam pipe or boiler due to 
tlie internal pressure force. In these 
cases the surfaces are cylindrical. 

Consider first a right circular 
cylinder (Fig. 151) placed with 
axis horizontal, and full of liquid. 
At any point of the surface the force 
is normal to the surface. Since the 
normal forces at various points on 




Fig. 154. 



140 



Elementary Mechanics 



this surface have different directions, the total pressure force as 
the sum of the normal forces has no physical significance. We 
can, however, find the total component of pressure force in any 
one direction, since the force vectors so involved are all parallel. 
Pass a vertical plane M through the axis of the cylinder. Then 
from symmetry, both of forces and areas, the sum of the hori- 
zontal components of the forces on the right-hand cylindrical 
surface is equal and opposite to that on the left-hand cylindrical 
surface, and these are both equal numerically to the total pressure 
force on each side of the vertical plane M. Hence the thrust T, 
to the right and left on the half cylindrical surfaces, is given by 
(area) X (w) X (depth of center of area), or 
T = 2rh .W'r = 2wrh lbs. 

Suppose the same cylinder is full of steam at a pressure of 

p pounds per square inch. Here 
the density is negligible, and so 
the normal force can be taken as 
constant in magnitude at all 
points of the inner surface. In 
this case the thrust on the verti- 
cal diametral plane is simply the 
area times the pressure, and so 

T = 2rh'P lbs., 

and this is constant for any 
diametral plane, vertical or other- 
wise. This force must be sustained by the material of the cylin- 
der, i. e., the two areas AC and BD (Fig. 155) are subjected to 
a stress of p' pounds per square inch, given by 
2plit = 2p]ir, 
'' p' — pr/t lbs. per sq. in. 

Applications of this result will be given in a later chapter 
(Art. 89). 




Fig. 155. 



Fluid Pressure 



141 



ProhJems 

235. What is the pressure 40 feet below the surface of a lake? 

236. A jar in the form of a frustum of a right circular cone, 
has a diameter of 14'' at the top and 10" at the bottom. It is 
filled with 9" of mercury and 8" of oil. If the specific gravity 
of oil is .915, and 1 cubic inch of mercury weighs -| pound, find 
the total force on the bottom of the jar. 

237. A 6' square is submerged vertically in salt water until 
its upper edge is 5' below the surface, and then it is turned about 
its horizontal upper edge through an angle of 60°. Find the 
force on the square. 

238. A triangle is submerged in liquid with its base in the 
surface. Where must a horizontal line be drawn across it, so 
that the forces due to the liquid pressure on the two parts shall 
be equal? 

239. If a full cylinder is held with axis horizontal, determine 
the resultant force acting vertically on the lower half of the 
curved surface. 

240. Find the total pressure force on the vertical face of a 
semi-elliptical dam 100' long and 15' deep. 

241. Find the force on a vertically submerged trapezoid of 
altitude 4' and bases 12' and 9', the long edge being in the surface 
of fresh water. 

242. If the pressure in the accumulator of a hydraulic forging 
press is 2 tons per square inch, 
and the diameter of the press 
cylinder is 80", what is the force 
of the press? 

243. Find the force on the face 
of a dam whose submerged verti- 
cal surface has the shape of an 
inverted cycloid of depth 2a. 

244. The pressure in a water 
pipe in the basement is 74.5 
pounds per square inch, while in 
the fifth story it is only 48 pounds 
per square inch. Find the dif- 
ference in elevation. 

245. Find AP so that the force 
on K will be three times that on 
E (Fig. 156). 




142 



Elementary Mechanics 



246. A board 2' wide at one end and 2'6" at the other, is 8' 
long. Find the force on the board when placed vertically in 
water with the narrow end in the surface. 

247. The back of a dam has a slope of 3 to 2. Find the hori- 
zontal force per linear foot upon it when the water is 13 feet deep. 

248. What head of water will burst a pipe of 24" internal 
diameter and f " thick, the tensile strength of the material being 
20,000 pounds per square inch ? 

249. An equilateral triangle is immersed vertically in water, 
with two vertices at a depth of 6' and the third at a depth of 9'. 
What is the force on it? 

80. Center of Pressure. — The forces producing pressure at 
points of a submerged plane area are all normal to the plane and 
so constitute a set of parallel forces. Their resultant is then a 
force parallel to these acting at the 
center of force. In hydrostatics this 
center of force is called the " center of 
pressure '^ (C. P.). 

Take any submerged plane area A 
(Fig. 157), making an angle a with 
the horizontal, and let I be the line of 
intersection of the plane of A and the 
free surface of the liquid. Measure y 
in the plane of A, from I to a repre- 
sentative area element dA, Then the 
head on this element is y sin a, and the 
element of force is ivy sin a-dA. Let Tfp be the distance from I to 
the center of pressure, measured in the plane of A. Then if F is 
the total pressure force on A, we have, taking moments about I, 




Fig. 157. 



yp-F= \y- {wysina-dA), 



the integration to cover the area. 



Fluid Pressure 143 

If a==0, the plane is horizontal, and there is no line I. Conse- 
quently this result is valid for a:fiO. But in the case of a hori- 
zontal plane, the center of pressure is obviously the center of area. 

If A has a line of symmetry, s, perpendicular to I, then both 
the center of pressure and the center of area lie on s, because of 
the equal forces and equal areas to right and left of s. The 
position in A of the center of pressure is then determined by yp 
and s. If there is no line of symmetry perpendicular to I, 
moments can be taken about any line perpendicular to I in A, 
giving the horizontal distance from such a line to the center of 
pressure. 

Excluding the horizontal case, we then have, 

- _ Ja wy^ sin a- dA _w sin a \a y^dA _Sa y^dA _ Ii _Ii 
'^^~ F ~ wsina \AydA ~ ~JJ^/dA ~ Af ~ 'S ' 

where li is the moment of inertia of the area A about I, yg is 
the distance from I in the plane of A to the center of area, and S 
is the statical moment of A about I. Since 



we get 






where ^^^ is the squared radius of gyration of the area A about 
the line I. If kg~ is the squared radius of gyration of A about a 
line through the center of area parallel to I, we have 

UQ Ug 

81. Examples 

1. Find the center of pressure on a circular bulkhead of radius 
2 feet, situated vertically with its center 12 feet below the surface 
of the water (Fig. 158). 



144 
Here 



Elementary Mechaxics 



yp- 



h'+y/- f+(i'2)^ 



ys 



13 



12j\ieet=:irr; 



i. e., the center of pressure is on a vertical diameter, 1 inch 
below the center of the circle 




Fig. 158. 



Fig. 159. 



2. Find the center of pressure on a semi-ellipse (Fig. 159), 
submerged vertically, with minor axis in the surface, the semi- 
axes being 2 feet and 3 feet. 
Here 



y^^% 



But 
hence 



yp 



yy--2xdy 

y-2xdy 
Jo 

a;^ = |(9-r), 

y-\/9 — y--dy 

Jo ^ 



y\/9-y'.dy 
The denominator integral is of the " power form " and the 



Fluid Pressure 



145 



numerator is easily integrated after making the snbstitutiou 
7/ = 3 sin 6. The result is 

16 



yp 



1.77 feet c1ot\ti on the major axis. 



82. General Illustrative Examples 

We shall now apply the principles that have been developed to 
various problems in fluid pressure. 

1. A vertical rectangular ma- 
sonry dam is 4 feet thick and 
weighs 140 pounds per cubic foot 
(Fig. 160). Determine its height;, 
that the pressure force exerted by a 
full head of fresh water against one 
side of it may make it fail by sliding, 
given the coefficient of friction, 
/x = .75; and also the height at which 
it will fail by turning over. 

Take a foot length of the dam. Its weight 
TF = 4(l)/i(140)=560/i lbs. 
The resistance offered to sliding is then 

i^' = .75]7 = 420/i lbs. 
The total pressure force is 

F = li{l) . {li/'2)iD = Ubli-/-^ lbs. 
We must have F — F', and so 




IS 



or 



/i. = 13.44 ft. 



To be on the point of overturning about 0, we must have the 
moments of F and IT' equal about 0. F acts at the center of pres- 
sure, which is at height /i/3 feet above 0. IF acts at the center of 



gravity^ 



or 



and so has an arm of 2 feet. Hence, 
125/r/4.(/?/3)=560/i(2), 

/i = 10.37 feet. 



10 



146 Elementary Mechanics 

2. The fin keel of a yacht is shown in Fig. 161. Find the total 
pressure force, the depth of the center of pressure, and the 



18' 


1 • 

6' 






\ 




y ' 


5 


\ 




y 


\ 




y i 


\ 








v 






^^ O - 







Fig. 161. 

moment of the force which tends to turn the keel about the line 
of intersection of its plane with the surface of the water. 

Here, using Simpson's Rule with li — l, we have the following 
tabulated computation : 

y X c xyc x(xyc) 

18 6 1 108 648 

17 7 *4 476 3333 



15 


8 


2 


240 


1920 


12 


9 


4 


432 


3888 


8 


10 


2 


160 


1600 


3 


11 


4 


132 


1452 





12 


1 









ll 



1548 12810 

:,F = w- ^ •^xyc= y (1548)rr516t^ = 32250 lbs. 

(Where w; = 62.5 for fresh water.) 

/S = ?^.| .^x-'yc^-^ (12840) =4280w = 267500 1b.-ft. 
a:> = 4280/516 = 8.294 feet below the surface. 



Fluid Pressure 



147 



3. An automatic water gate may be used for draining a marsh, 
and is so constructed (Fig. 162) that it opens when tlie head of 
water on the marsh side exceeds that on the ocean side, and closes 



^ 



- -Marsh 



Oceari 



Si 

Fig. 162. 



when the condition is reversed by the rising of the tide. The dif- 
ference in the head, when closed, causes horizontal forces to act on 
the gate, at the contacts and H. These two forces, together 
with the two pressure forces, form a system of four parallel forces 
in equilibrium (Fig. 163). Taking moments 
about at the sill, we get the reaction R at 
the hinge H. What is the reaction, for a gate R 

4 feet high and 5 feet wide, when the ocean 
stands 3 feet above the sill, and the marsh 
(salt) stands 1 foot above the sill? Taking 
moments about gives 

P,(l/3)+7?(4)=P,.(3/3), 

where L ^^ 

P,= (l) (5) (64) (1/2) lbs. __R_ 

(the density of salt water being 64 pounds O 

per cubic foot), and Fig. 163. 

P,= (3) (5) (64) (3/2) lbs., 
from which 

i? = 346f lbs. 



148 



Elementary Mechanics 



Prohleins 

250. Find the C. P. of a rectangle submerged vertically, with 
its edge in the surface. 

251. Find the C. P. of a vertically submerged triangle when 
(a) one side is in the surface, and (b) one side is parallel to the 
surface and the opposite vertex is in the surface. 

252. Find the C. P. of a vertically submerged circular quad- 
rant with one edge in the surface. 

253. Find depth of C. P. of a vertically submerged semi- 
ellipse having its major axis in the surface. 

254. A parabolic segment of base 26 and height h is submerged 
vertically. Find the C. P. when (a) its base is in the surface, 
and (b) its base is parallel to the surface and its vertex is in the 
surface. 




14 

Fig. 164. 



Fig. 165. 



255. Find the C. P. of a submerged trapezoid, having bases 
a and b and altitude h, if base a is in the surface. 

256. A pair of flood gates are swung on hinges 3/2' from top 
and bottom of the gates which close an opening 10' wide by 6' 
deep. What is the force on each hinge when fresh water stands 
at 'the top of the gate? 

257. Find the C. P. of a vertically submerged circle, if the 
head on its center is 2 diameters. 

258. If the head on one side of a tide gate is 7' and on the 
other 4', find the resultant force and its point of application. 

259. A vertical masonry dam is 4' thick, 18' high, and weighs 
125 pounds per cubic foot. How high can fresh water stand 
against the dam before it overturns? 



Fluid Pressure 



149 



260. A cubical tank is 3' on an edge. One vertical face is 
hinged at the top. The tank is filled with fresh water. What 
is the least force which will keep the liinged face in place, and 
where should it be applied? 

261. The partition wall between the compartments of a set- 
tling basin is 12' high. Water is within 1' of the top on one side, 
the other side being empty. If the masonry weighs 160 pounds 
per cubic foot and 60 per cent of its total weight represents its 
ability to resist sliding, how thick should the wall be? 

262. Find .T so that the force on P is 50 per cent greater than 
that on Q (Fig. 164). 

263. Triangle and rectangle are vertically 
submerged. JFind the altitude of the triangle 
so that the force on it shall be 50 per cent 
greater than that on the rectangle (Fig. 
165). 

264.^ A dam whose cross section is a right 
triangle of base 4' and altitude 15' weighs 
125 pounds per cubic foot. How high can 
salt water stand against the vertical face 
before it will turn over ? 

265. A rudder in the form of a semi-para- 
bolic segment (see Fig. 166) has AB verti- 
cal, and A Sit a, depth of 3'. Find the total 
pressure force and its point of application. 

Review Prohlems 

266. Find force and C. P. on a vertically submerged circle 
tangent to the surface. 

267. Find force and C. P. on trapezoid with bases 10' and 8' 
and altitude 3' submerged vertically in fresh water with 10' 
edge in the surface. 

268. Find the moment of the force on a vertically submerged 
circle tangent to the surface, about the tangent line in the surface. 

269. Find force and C. P. on a rectangle 6' x 9' submerged 
vertically in salt water, having short side horizontal and upper 
one 5' below the surface. 

270. If the centers of two circular pistons, of 4" and 12" 
diameter are at the same level, and the smaller one is pushed 




Fig. 166. 



150 



Elementary Mechanics 



inwards with a force of 100 pounds, how much force must be ap- 
plied to the larger one to keep it from moving? 

271. What is the pressure 1000 fathoms deep in the sea? 

272. A cubical vessel is filled with liquid. Compare the forces 
on its bottom and sides. 

273. A vertical water gate 40' wide has fresh water standing 
20' deep on one side. How deep must salt water stand on the 
other side, in order that the' forces may be equal ? 

274. Find the C. P. for a rectangle axb vertically submerged, 
with one vertex in the surface, and a diagonal horizontal. 

275. On one side of a sluice gate, water stands 8' deep, and on 
the other side 4' deep. The gate is 6' wide and 10' deep. What 
is the force on the hino-es, and where is the C. P. ? 





-J 












/ 




1 

8 


X 


p 








io' / 

/ < 


14' \ 
20' \ 




r 

Fig 


r. 167. 




Fi 


G. 168. 



276. Find the magnitude and point of application of the force 
on a vertically submerged rectangle 4' x 6' with short side in the 
surface. 

277. A solid rectangular masonry dam is 6' thick and weighs 
136 pounds per cubic foot. Find its height in order that it will 
just be on the point of overturning, when fresh water stands 
against one face at a depth of | the height of the dam. 

278. A dam whose cross section is a triangle has a vertical 
back, is 3' wide at the base and 15' high. Find the height to 
which the water may rise behind it, in order to cause failure 
(a) by sliding, and (b) by turning, using 0.75 as the coefficient 
of friction and 140 pounds per cubic foot for weight of masonry. 



Fluid Pressure 



151 



gate 5 



279. What is the force on the hinges of an automatic water 
high and 8' wide, when the sea stands -i' above the sill 

on one side, and salt marsh water 2' above the sill on the other 
side? 

280. Find the magnitude and point of application of the force 
on a vertically submerged triangle of base 2' and altitude 3' when 

(a) the vertex is in the surface and the base is horizontal, and 

(b) the base is in the surface. 




281. Find x so that the force on P equals that on Q (Fig. 167). 

282. Find the force on this vertically submerged trapezoid 
(Fig. 168). 

283 A solid rectangular masonry dam weighing 110 pounds 
per cubic foot is 12' deep. Find the width, if the dam is to be 
60 per cent wider than the width at which it would just overturn 
imder a head of 10' on one face. 

281. Find magnitude and point of application of the force on 
the area in Fig. 169 if submerged vertically in fresh water. 

285. What depth of water in a cylindrical jar of radius r 
produces the same force on the sides as on the bottom ? 



152 



Elementary Mechanics 



CHAPTEE YII 

DEFORMABLE BODIES 

83. Introduction. — The members of any machine or structure 
are acted upon by forces and transmit forces. In Chapter IV we 
studied the forces acting upon and transmitted by the various 
members of a frame, and we can study with the aid of kinetics the 
forces acting upon and transmitted by the moving members of a 
machine. In both cases we assume that the members are rigid 
bodies, that is, remain unaltered regardless of the magnitude of 
the forces acting upon them. Experience, however, tells us that 
actually there is a limit to the magnitude of the forces which can 
be applied to any member with safety, and that this limit is re- 
lated to the shape, dimensions and material of the member. In 
this chapter we consider the behavior of material under the action 
of force. This study is sometimes called " Strength of Material.^^ 
In treating such problems it is necessary to use results obtained 
by experiment with actual materials. Tables giving these results 
may be found in any handbook and a brief table follows : 





wt. lbs. 

per 
cu. ft. 


Ultimate strength 
in lbs. per sq. in. 


Elastic 
limit in. 
lbs. per 
sq. in. 


Modulus 
in lbs. 

per 
sq. in. 


Factors of 
safety 


Material 


Ten- 
sion 


Com- 
pres- 
sion 


Shear 


steady 
load 


Shocks 


Struct, steel . . . 

Cast iron 

Wrought iron. . 
Yellow pine. . . . 
Oak . . . 


490 
450 
480 
40 
48 
125 
150 


60000 

22500 

50000 

9000 

10000 

500 

300 


60000 
90000 
40000 
7000 
6000 
3000 
2500 


50000 
20000 
40000 
1500 
4000 
1000 
1000 


35000 
GOOO 

25000 
30C0 
3000 


30 X 10« 

15 X 10« 

28 X 10« 

150 X 10^ 

150 X lO'' 


4 to 10 
8 to 20 
4 to 10 
8 to 15 
8 to 15 


Brick 


15 to 40 


Concrete 


1000 


2000 X 10^ 


15 t 


40 



Defor:\iable Bodies 153 

84. Normal Stress. — A rod acted upon by two equal and oppo- 
site forces P (Fig. ITO), applied to its ends in the direction of 



-»P 



B 

Fig. 170. 

its axis^ is said to be subject to a '' Xormal Stress " and the force 
P is called the " Load." If we consider a cross section of the bar 
by a plane perpendicular to its axis, equal and opposite forces, 
P, must act on either side of the section to preserve equilibrium. 
The force P is assumed to be distributed uniformly over the 
cross-sectional area. The load per unit cross-sectional area is 
called the " Stress " and we shall denote it by p. Thus, 

Stress = Load -^ Cross Section. 

If A denote the cross section, p — P/A. 

The stress is called " Tensile Stress '' and the bar is under 
tension when the two forces P tend to sej^arate the parts on the 
two sides of the section, and is called " Compressive Stress," and 
the bar is under compression when the forces tend to push the 
parts together. Tensile stresses and compressive stresses are 
called " Xormal Stresses " since they act at right angles to the 
section of the bar. 

Example. — A hollow cast-iron column of six inches exterior 
and five inches interior diameter sustains a load of 50,000 pounds. 
Find the compressive stress due to this load. 

Solution. — 

7r(36 — 25)/4 = area of cross section of metal in square inches. 
P = 50,000 lbs. 

.•.2?=P/A = 50,000--ll7r/-i = 50,000 -tr- iV 
= 5,800 lbs. per sq. in. (Approx ) 



154 



Elementary Mechanics 



Problems 

286. A wrought-iron rod one inch in diameter is nnder a 
tension of 8000 pounds. Find the stress. 

287. The diameter of the piston head of a steam engine is 
16 inches and that of the piston rod 2^ inches. If the engine is 
working nnder a steam pressure of 100 pounds per square inch, 
what is the maximum stress in the rod ? 

288. The head of an engine cyhnder is 10 inches in diameter 
and the steam pressure applied to it is 125 pounds per square 
inch. The head is fastened on by 8 steel bolts, J inch in diameter. 
Find the tensile stress in each bolt. 

289. Calculate the size of a square wrought-iron bar to stand 
a pull of 3000 pounds without breaking. 

85. Longitudinal Strain, — If a bar is stressed in the way de- 



70,000 
60,000 
























50,000 
40,000 
30,000 
20,000 
10,000 

























^^ 




















r 


































































1 























0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20 0.22 

Unit Elongation — Change of Length Per Unit of Length. 
Fig. 171. — Stress-Strain Diagram. 



scribed above a change in length takes place. It is usual to con- 
sider the change in length per unit length and to call this the 



Deformable Bodies 155 

'• strain/^ which we shall denote by e. Thus, if a bar I inches long 
is elongated or contracted x inches, e — x/l. 

86. Stress-Strain Diagram. — If a test piece of material is placed 
in a testing machine and the load slowly applied, we find that the 
deformation of the piece is gradual and proportional to the stress 
until the stress reaches a certain value. This value is called 
the " Elastic Limit,^^ since, if the stress has not exceeded it, the 
piece will resume approximately its original shape when the load 
is removed. As the stress increases beyond the elastic limit the 
deformation increases more rapidly until the stress reaches a 
value when again there is a marked difference in the way the 
deformation changes. This second value of the stress is called 
the " Yield Point." From the yield point on, the stretch is very 
rapid with little change of stress until finally the piece breaks. 
The value of i\\Q stress when the rupture takes place is called the 
" Ultimate Strength." 

As might be supposed, materials vary widely in their ability 
to withstand different kinds of loads. Steel is about equally 
strong in tensitm and compression. On the other hand, concrete, 
while strong in compression, can stand little tension and hence 
must be reinforced when it is to carry a tensile load. 

87. Hooke's Law — Modulus of Elasticity. — Hooke's Law 
states that within the elastic limit the ratio of stress to strain 
is constant. This ratio is called the " Modulus of Elasticity " 
or " Young's Modulus " and is represented by E. Thus, E — ^/e. 
As stated in the preceding paragraph, the variables p and e are 
connected by a straight-line law. Since e is merely the ratio 
of two lengths it is independent of the units of measurement, 
and hence E is always measured in the same units as y, usually 
pounds and square inches. The value of E varies for different 
kinds of materials but not with the shape or size of the piece. 



15G Elementary Mechanics 

Example. — A bar ol' stool, 10 l'oi>l lono- and 3 squnro inches in 
section, stretches -.y^ inclios undor a loaci of GO, 01)0 i)oiiiul8. Find 

The stress, p=:G0,000/3 lbs. per sq. in. 

The strain, g--^^^\20. 

'Vhc modulus, ii'=:/;/e = 30,000,000 lbs. per sq. in. 

290. How mnoh is a 100-l'oot slool (ape, -^ inch wide and -/g- 
inch thick, lengtliened by a juill ol* 40 ])ounds? 

291. A wronght-iron tie rod, 1|: inches in diameter, is slri^lchod 
■} inch by a pnll of 18,000 pounds. Find the original length of 
the rod. 

292. If the breaking strength of a wiiH^ roju^ is 90,000 pounds 
per sqnare inch, lind the greatest load which an elevator ro])e ^ 
inch in diameter can sustain. 

293. If a steel shaft is suspended vertically from one end, 
how long will it have to be in order to break under its own weight? 

294. Find thc^ modulus of elasticity ot a copper wire 2G feet 
long and 0.09 inches in diameter if it sii-(4clu>s 0. 1 inches under 
a })ull of 120 pounds. 

88. Working- Stress and Factor of Safety. — In engineering 
practice the allowable stress in any maicn-ial must be well within 
the elastic limit, for beyond tliat point the stress causes a per- 
manent s(>l- in 11)0 material and it will not even approximately 
assunu=i its original shape when relieved of the h)ad. Tin's allow- 
able stress is called the "Working Stress" and is the stress on 
which the design of the niond)er depends. It is obtained by 
dividing ilio ultinuUe strength by a constant called the "Factor 
of Safety." The choice of this factor depends on many things, 
among which are the uniformity of the material, whether the 
loads are gradually oi- suddiMily applied, clc. Tlu> choice of the 
]n-oper factor is largely a matter of experience, and, for various 
materials and various kinds of loads, it runs from 4 to 35 or 40. 



Deform ABLE Bodies 157 

Prohlems 

295. Find the factor of safety in Probs. 286, 287, 288. 

296. A short wooden (oak) column, 10 inches in diameter, 
sustains a load of 9500 pounds. Find the factor of safety. How 
great a load could this column sustain if the factor of safety 
were 8 ? 

89. Shear. — We have seen that the normal stress acts at right 
angles to the section of the bar and tends to move the molecules 
in that direction. It often happens that the member is so 
loaded that the molecules tend to move past each other in a 
direction parallel to the section. Such an action is called a 
" Shear " and the load a " Shearing Force." The shearing force 
per unit of area is called " Shearing " or " Tangential Stress," 
is denoted by q, and is measured in pounds per square inch. Thus, 
if F is the shearing force and A the cross-sectional area, q — F/A. 

Two plates bearing a longitudinal load and held together by 
a rivet present a good illustration of shearing stress on the rivet 
(Mg. 172). 

If the rivet fits the holes closely and the load is sufficiently 




F^ 



Fig. 172. Fig. 173. 

great, it will be sheared off smoothly as though a plane had been 
passed perpendicular to its axis. 

If a bolt is in tension a shearing force acts on its head, tend- 
ing to strip it from the bolt (Fig. 173). The shearing stress 
here acts along the cylindrical surface which is the continuation 
of the surface of the shank. 

Example 1. — A steel bolt 1 inch in diameter with a head IJ 
inches deep is subjected to a tension of 15,000 pounds. Let it 
be required to find the shearing stress in the head. 



158 Elementaky Mechanics 

The area over which the shear acts is the lateral surface of a 
right circular cylinder which equals tt- 1 • 5/4 = 3.93 square 
inches. Then ^ = 15,000/3.93 = 3,817 pounds per square inch. 
Taking the ultimate strength in shear to be 50,000 pounds per 
square inch, the factor of safety against shearing is 50,000/3,817 
= 13.1. 

Example 2. — A boiler, 48 inches in diameter, is made of -^ 
inch plate joined by a single-riveted lap joint. The rivets are 
f inches in diameter and the pitch is 2 inches. If the internal 
steam pressure is 120 pounds per square inch, what will be the 
shearing stress in the rivets? 

A single-riveted lap joint is formed by overlapping plates 
held in position by a single row of rivets. The pitch is the dis- 
tance between the centers of the rivets. 

From Art. 79 the force on each rivet is P — p'rh, where p' is 
the steam pressure, r the radius of the boiler in inches and h 
is the pitch. 

Thus, 

P = 120 . 24 . 2 = 5760 lbs. per sq. in. 
Then 

^ = P/^ = 5760- ^^ =13,032 lbs. per sq. in. 

If there had been two rows of rivets, called a double-riveted lap 
joint, the stress in each rivet would be half what it was in the 
first case. A more complete discussion of riveted joints can be 
found in a handbook. 

Problems 

297. Compute the shearing stress in the heads of the bolts in 
Problem 288, if the heads are f inches deep. Find also the 
factor of safety against shearing. 

298. Compute the shearing stress on the wrought-iron rivets 
of a boiler 3 feet in diameter, which is carrying a steam pressure 
of 120 pounds per square inch, if the joint is a double-riveted 
lap joint with | inch rivets and 1:^ inch pitch. Find factor of 
safety against shear. 



Deformable Bodies 159 

90. Equal Strength. — Suppose it is required to find the thick- 
ness of the head of a steel bolt 1 inch in diameter so that the bolt 
will be equally strong in tension and shear. 

The tensile load that the bolt will carry will be tt • 1^ . 60,000/4. 
The total shear (see Fig. 173) which the head will carry will be 
TT'l't' 50,000 where t is the thickness of the head. Since these 
loads must be equal 

TT . P . 60,000/4 = 7r . 1 . ^ 50,000. 
Therefore, 

, 60000 o/m- 1. 
^= 1750000 ='/^^^^^^^^- 

If a riveted joint, as that of a boiler, fails because the plate 
ruptures, the break will occur along the outer row of rivets, since 
the force required to produce the break is least here. Eivets 
lessen the strength of the plate, since metal is removed for the 
holes. Hence the strength of the plate must be taken at its 
weakest section, i. e., along a row of rivets. 

Example. — A boiler plate is -| inch thick and there is a double- 
riveted lap joint of f inch rivets ; what must be the pitch of the 
rivets for equal strength of the joint in tension and in shear? 

Since the material of the rivets and plate is the same, the 
same factor of safety would be used for tension and shear, and in 
the numerical work would cancel out. Hence instead of the work- 
ing stresses the ultimate strengths in tension and shear, 60,000 
pounds and 50,000 pounds per square inch, respectively, can be 
used. 

The area of a section of the plate between two consecutive rivet 
holes will be i(p — |) where p is the pitch of the rivets. The 
strength of the plate is then 

60,000. i.(p-l). 

Since there are two rows of rivets, the total load will be borne 
by two rivets whose strength in shear will be 



160 Elementary Mechanics 



whence 



2. 50,000. TT. 9/(4. 16), 
60,000.1. (p-|) =2. 50,000. 77. 9/(4. 16), 

p = 2.22 inches. 



Problems 

299. If the boiler in Prob. 298 should explode, wonld the 
failure be due to the shearing of the rivets or to the rupture of 
the steel plate (assume the plate {q inch thick) ? 

300. The efficiency of a joint is its minimum strength divided 
by the strength of the unpunched plate. Find the efficiency of 
the joints in'Prob. 298. 

301. If the thickness of a boiler is t, the diameter of the rivets 
d and the pitch of the rivets p, find the relation between d and t 
that will make the joint equally strong in tension and in shear, 
assuming a double-riveted lap joint. 

302. A bolt 1|" in diameter has a head J" thick which shears 
off under a load of 12,000 pounds. Find the ultimate shearing 
strength of the bolt material. 

303. Two \" plates are double-riveted with single butt-strap 
joint, the rivets being |" in diameter. If q — ^p, find the pitch 
for equal strength. 

304. A single-riveted lap joint, plate %" thick, 8" wide, con- 
tains 3 rivets, each J" in diameter. If the plate is subjected to 
a pull of 12,000 pounds, find the tensile stress of the plate, the 
shearing stress on the rivets, and the efficiency of the joint. 

305. Determine the thickness of head for 1^" wrought-iron 
bolt, if the tensile strength of the bolt is to jje equal to the 
strength of the head against shearing. 

91. Beams. — If a horizontal bar is under the action of forces 
perpendicular to its axis, it is called a beam. The forces which 
act upon the beam may be either loads concentrated at points or 
distributed over the beam, this distribution being uniform or 
varying in any manner. Beams are classified according to the 
manner in which they are supported. A beam may simply rest 



Deform ABLE Bodies 161 

on any one of its supports, in which case it is said to be supported 
at this point; or it may be held rigidly fixed at the support, in 
which case it is said to be fixed at the point. A beam supported 
at its ends (Fig. a) is called a simple beam; a beam fixed at both 



i 



— ^ 



^ 
^ 



m — zs — 6 — zs — m 

^=1 



zs — 

Fig. 174. 

ends is called a huilt-in beam (Fig. b) ; a beam free, i. e., not 
supported at all, at one end and fixed at the other is called a 
cantilever beam (Fig. c) ; a beam extending over more than two 
supports is called a continuous beam (Fig. d) ; a beam over- 
iianging its supports is called an overhanging beam (Fig. e). 
11 



162 



Elementary Mechanics 



92. Kinds of Stresses in Beams. — If a cantilever beam is 
loaded at its outer end, the load tends to rupture the beam by 
bending as shown in Fig. 175. 




W 



Fig. 175. 




If, on the other hand, a concentrated load is applied close to 
one of the supports of a simple beam, the load tends to rupture 
the beam by shearmg ; that is, to make the parts of the beam 



W 



7^ 




Fig. 176. 



separate by slipping past each other in a vertical plane (Fig. 
17G). These are the two principal kinds of stresses, or tenden- 
cies to rupture, occurring in beams. In general at any point of 



Deformable Bodies 163 

a beam both kinds are found. In order that the beam may sup- 
port the required load it must be strong enough to resist both of 
them. The following sections will deal with the determination 
of the magnitude of these stresses due to any kind of loading. 

93. Shearing Force. — If a beam with any kind of loading and 
supporting forces, for example those in Fig. 177, is thought of as 



Q^> ' 1 

qis 1 



Fig. 177. 

cut at any point by a plane HH, each part of the beam must be 
in equilibrium under the action of all the forces which act on it. 
Now just as in applying the method of sections to frames (Art. 
51) these forces are of two kinds, the external forces and those 
due to the stress at the plane of section, i. e., the force exerted by 
one part of the beam on the other. (Note that the word " stress " 
as used in this chapter refers always to intensity of stress, not 
to total force as in Chap. IV.) Thus the vertical forces acting 
on the left-hand part of the beam in the figure are the two loads 
TF-L and TFo, the supporting force R^, and Q, the shearing force 
at the section HH. As all of these but the last are known or 
readily found, we can find the last also by applying one of the 
conditions for equilibrium, namely, by setting 2F = 0, Thus 
the shearing force at any section is numerically equal to the 
algebraic sum of all the vertical forces acting on one side of the 
section. 



164 



Elementary Mechanics 



94. Algebraic Sign of the Shearing Force. — As the shearing 
force always tends to move one part of the beam up and the other 
down, it is necessary to establish a definite convention for. its 
algebraic sign. The shearing force is called positive at any sec- 
tion if the tendency of the external forces is to move the part to 
the left of the section np and that to the right down, negative in 
the contrary case. 

The student should note carefully that generally ihe shearing 
force varies with the position of the section and will be constant 
for all sections of the beam only in one special case. 

95. Shearing-Force Diagrams. — It is often necessary to find 
the shearing force for every section of a beam. To facilitate this 
calculation we plot the value of the shearing force as an ordinate 
vertically above the section for which it has been calculated. 
The locus of these points is a curve or diagram. The origin is 
taken at the left end of the beam, and the units on the horizontal 
axis represent units of length, while those on the vertical axis 
represent units of force. 

Example 1. — Draw the S. F. diagram for a simple beam, 12 



200ibs 

2 4^ 3 v^ 



30Qi 



SOOibs 



R^ aooibs 



Fig. 178. 



feet long, loaded as sliown in Fig. 178, the weight of the beam 
being neglected. 

The reactions R-^ and R^ are found by taking moments about 
the supports and using as a check that R^ + R.-, = svin\ of the loads. 
We start at the left reaction and lay off 7^^ = 300 to scale, using 
any convenient unit (Fig. 179). Since there are no forces 



Deformable Bodies 



165 



between R^ and the 200-ponnd load, the S. F. must have the 
same value throughout this interval and the diagram is the line 
ah. When we pass over the 200-pound load the S. F. drops to 
100 pounds, since the resultant force to the left of this section 
is 300 — 200 = 100 pounds. The diagram falls to c and continues 
horizontal until the 100-pound load is reached at d, and then 
drops to e. The S. F. at this section and any section beyond up 
to the 300-pound load is zero, and the curve ef lies along the 
beam. From / the diagram drops to g (—300) by passing over 
the load of 300 pounds. It continues in the horizontal direction 
to li, where the value is — 300, or the negative of the reaction R^. 

SF 




Fig. 179. 



The S. F. at the supports is always numerically equal to the re- 
actions. The complete S. F. diagram then is the broken line 
abcdefgh. If we denote the shearing force at any section x 
units from the left end by Q{x), we may write the equations 
for the S. F. curve as follows : 

Q{x) =300 for X between and 2 feet; (J (a;) =100 for x be- 
tween 2 feet and 5 feet; Q{x) =0 for x between 5 feet and 9 feet; 
Q{x) = -300 for x between 9 feet and 12 feet. 

Example 2. — Draw the S. F. diagram for a simple beam, of 
length I feet, uniformly loaded with iv pounds per foot run. 

The total load will be ivl pounds and as the loading is sym- 
metrical the reactions are each wl/2 pounds. Now if we take 
any section AB, at a distance x from the left support, the S. F. 
will be iul/2 — iux. Thus the equation of the curve is 

Q{x) =wl/2 — wx. 



166 



Elementary Mechanics 



This is the equation of a straight line and the diagram is the 
line acl shown in Fig. 180. The S. F. is zero when x = \l. 




-— X 



Fig. 180. 



Problems 

306. Construct the S, F. diagram for a cantilever beam of 
length 15 feet with a uniformly distributed load of 250 pounds. 
What is the S. F. 3 feet from the fixed end? 10 feet from the 
fixed end? 

307. Draw the S. F. diagram for a simple beam 20 feet long 
with a concentrated load of 1000 pounds at the center and 
another of 540 pounds 3 feet from the right end. What is the 
S. F. 2 feet from the left end? 12 feet from the left end? 
1 foot from the right end ? 

308. Draw the S. F. diagram for a cantilever beam of length 
12 feet with a uniform load of 40 pounds per foot run and a con- 
centrated load of 2000 pounds 3 feet from the free end. 

309. A simple beam 16 feet long carries a uniform load of 40 
pounds per linear foot, and two concentrated loads, 240 pounds 
3 feet from the left end and 180 pounds 4 feet from the right 
end. Draw the S. F. diagram and find the S. F. 4 feet from 
each end. 

310. A simple beam 20 feet long carries 2000 pounds 5 feet 
from the left end, 5000 pounds 4 feet from the right end, and in 
addition a load of 200 pounds per foot run; find the S. F. at 
the ends and center and draw the S. F. diagram. 

311. A beam 20 feet long, supported at its ends, carries 8 tons 
6 feet from the left end, 3 tons 9 feet from the left end, and 2 



Deformable Bodies 



167 



tons 14 feet from the left end. Draw the S. F. curve and give 
value of S- F. at the center. 

312. A beam 10 feet long has its left end fixed in a wall, and 
it carries a uniform load of 80 pounds per foot run, together 
with a concentrated load of 600 pounds 7 feet from the wall. 
Draw the S. F. curve and give the value of the S. F. 8J feet from 
the wall. 

96. Bending Moment. — In considering the shearing force in 
a beam we applied only one condition of equilibrium (2t/ = 0) 
to the part of the beam on one side of the section. From the 
definition of a beam, the external loads have no horizontal com- 



\ 






Ws W4 


^ 




R. 








/ 


F 



Rz 



Fig. 181. 



ponent. So the horizontal components of the forces due to the 
internal stresses must amount to zero at any section (Fig. 181). 
It follows from this that at any section where there is normal 
stress across the section, it must be tensile in one part of the 
plane of section and compressive in the other. That is to say, 
these forces form a couple. 

Before considering the exact distribution over the section of 
these normal forces, we can determine the magnitude of the 
Qouple they form by applying the condition of equilibrium 
21f=r0 to the part of the beam on one side of the section and 
taking moments about an axis in the section in order to elimi- 
nate the shearing force at the section. 



168 



Elementary Mechanics 



Example. — A simple beam 10 feet long is loaded with weights 
of 3000, 2000, and 6000 pounds at points 2, 4, and 8 feet 
respectively from the left end. Find the magnitude of the bend- 
ing moment at a point 6 feet from the left end. 

The supporting forces 4800 and 6200 are found by the ordi- 
nary method of statics applied to the beam as a whole (Fig. 182) . 
Considering now the part to the left of the section Till, the 
moments of the external forces about a point in //// will be 
4800 • 6 or 28800 clockwise and 3000 • 4 + 2000 • 2 = 16000 counter- 
clockwise. That is, the external forces exert a total moment of 



6000 



3000 H 
2000 1 



I 20C 

'4- 2'4- 



4800 



Fio. 182. 



6200 



28800-16000 or 12800 pound-feet clockwise. But since the 
left-hand part of the beam is in equilibrium the internal forces 
at the section must furnish a moment of 12800 pound-feet 
counter-clockwise. 

The principle exemplified here is sometimes expressed in the 
statement that the bending moment is equal to the resisting 
moment, the bending moment being the algebraic sum of the 
moments of the external forces on the part of the beam on one 
side of the section, the resisting moment being that of the force, 
exerted on this part of the beam by the other part, and trans- 
mitted by the internal stresses at the section. 

The student should notice tliat, like the shearing force, the 
bending moment will vary from point to point in the beam. 



Deformable Bodies 169 

. 97. Sign of the Bending Moment. — As with the shearing force, 
a definite convention for the sign of the bending moment is also 
necessary, since it is obviously clockwise on one side of the sec- 
tion and counter-clockwise on the other. The bending moment 
is called positive at any point when the external forces tend to 
turn clockivise the part of the beam to the left of the section, 
and of course negative in the contrary case. The reason for this 
convention is that when the moments act in the way we call 
positive they tend to bend the beam into a curve which is con- 
cave upward, that is, has positive curvature. It may be worth 
noting that if we regard it from this point of view we can see 
that the bending moment retains its sign even if we look at the 
beam from the opposite side, thus interchanging left and right. 
This is not the case with the sign of the shearing force. 

Example. — A simple beam 12 feet long has a uniformly dis- 
tributed load of 300 pounds per foot run on the left half. Find 
the bending moments at 4 -feet and 8 feet from the left end. 



W///////////////////////M 


I'll 


V A B 

R. 

Fig. 183. 


T 

'r 



Solution. — To find the supporting forces we may regard the 
load as concentrated at its center of gravity. This gives Rj^ = 
1350 pounds and i?. = 450 pounds (Fig. 183). 

The bending moment at A, the first point, is 4:R^ clockwise 
and the moment (counter-clockwise) of the part of the load to 
the left of A. This latter is 4-300 or 1200 pounds acting (on 
an average) at a distance of 2 feet from A. Hence the bending 
moment at A is 47i^, -2 • 1200 = 5400-2400 = 3000 pound-feet. 
We may find the bending moment at B in the same way ; that is, 

8.1350-5.1800 = 10800-9000 = 1800 Ib.-ft. 



170 Elementary Mechanics 

It is easier, however, to find it as the total count er-cloclcwise 
moment to the right of the section, i.e., 4i^2 = 4-450 = 1800 
pound-feet. 

98. Bending'-Moment Diagram. — Since the bending moment 
varies from point to point as we pass from one end of the beam 
to the other, the easiest way to get the general idea of its behavior 
is to draw its graph regarded as a function of x, the distanct? 
from one end of the beam. 

Example 1. — A simple beam 10 feet long carries loads of 800 
and 600 pounds at distances of 2 feet and 6 feet respectively 
from the left end. Draw the bending-moment diagram. 

Solution. — The supporting forces are 880 and 520 pounds at 
the left and right ends respectively. 



800 ^^^ 

600 



T 



R2=520 
R,=880 

Fig. 184. 



The length of the beam is divided into three intervals by the 
two concentrated loads. In the first of these if we take tlie 
bending moment at any point distant x feet from the left end, 
the only external force to the left is the reaction R^. The 
bending moment M^ then is 880a;. We must be careful to note 
that this formula holds only until we reach the first load. After 
that the moment of this load must be brought in, giving 

7lfo = 880a;-800(.T-2)==80.T + 1600. 

Similarly in the third interval, 

ilf3 = 880.T-800(a;-2) -600(.T-6), 
= -520^ + 5200. 



Deformable Bodies 



171 



We have a check here, as ^1/3 = when a: = 10, which we can 
easily see is correct by considering the right-hand part. The 




x=o 



graph (Fig. 185) then is determined as follows: 



i¥, = 880a; 


0g:ri2 


(1) 


M2 = 80a;+1600 


2<x<6 


(2) 


ilf3=-520a: + 5200 


6^.T^10 


(3) 



It is important for the student to note that a formnla whicli 
holds for one interval is entirely incorrect in any other. Two 
consecutive formulas, however, agree at their common end point. 
Thus (1) and (2) both give 1760 when x = 2 and (2) and (3) 
both give 2080 when x = 6 

In this problem each of the formulas for M is the first degree 
in X, hence the graph in each interval is a straight line. It is 
easy to see that this will always be so for concentrated loads 
since the distance x enters only in the arm of the moment. 
Hence in order to draw the B. M. curve we need compute the 
B. 'M. only at the points where the loads are applied, plot these 
and connect them by straight lines. 

Example 2. — Draw the B. M. curve for the example of xlrt. 97 
and find the maximum bending moment in the beam. 

Solution. — In the first interval, 

0ST^6, M^ = 1350x-x/2. (SOOx), 
-13d0x-150x'. 
For the rest of the beam 

6<'2:gl2, il/2 = 1350.T-1800(.T-3)=: -4500; + 5400. 



172 



Elementary Mechanics 



To find the maximum value of M^, we set the derivative of 
1350a; -150.T" equal to zero, 1350- 300a; = or x-4:i. There 
is no maximum in the second interval, as M is there always de- 



x = o 




4.5 6 

Fig. 186. 



creasing. When x = 4:^, M = 30o7i, the maximum B. M. (Fig. 
186). 

Problems 

313. A simple beam 20 feet long has loads of 300, 450 and 
250 pounds placed at distances of 5, 11, and 17 feet respectively 
from the left end. Find the B. M. and S. F. at 6, 10, and 15 
feet from the left end. 

314. A simple beam 29 feet long carries a load of 80 pounds 
per foot run. Find the B. M. and S. F. at 10, 19, and 26 feet 
from the left end. 

315. If the beam in Prob. 314 also carries a concentrated 
load of 1000 pounds at its midpoint, find the B. M. and S. F. 
for the same points as in Prob. 314. 

316. Draw the B. M. and S. F. diagram's for the problems in 
307, 308, and 309. 

317. A beam 8 feet long is fixed at the left end and carries a 
concentrated load of 300 pounds 3 feet from the right end. 
Draw the B. M. and S. F. diagrams. 

318. A beam, 12 feet long, supported at its ends, carries con- 
centrated loads of 300 and 700 pounds at points 4 and 9 feet 
respectively from the left end. Draw the B. M. and S. F. 
diagrams. 

319. A 12-foot beam, supported at the ends, carries 6 tons 8 
feet from the left end, and a continuous load of f ton per foot- 
run. Draw the S. F. and B. M. curves and find the B. M. 10 
feet from the left end. 



Deformable Bodies 



173 



99. Relations Connecting Loading, Shearing Force and Bend- 
ing Moment. — We have seen in the examples of the preceding- 
section from the diagrams of Q and M that where there is a 
concentrated load there is a discontinuity in the value of ^. In M 
itself there is no discontinuity, hut there is in its slope. Also 
there is a discontinuity in the slope of Q, where there is a 
sudden change in the rate of loading in the case of a distributed 
load. At other points, however, we found that these graphs 
were straight lines or smooth curves. We shall now derive two 
very simple relations connecting M, Q, and L, the rate of loading 
in a distributed load. 



X 

Fig. 187. 



Let Q and Q-\-^Q be the values of the shearing force at two 
points distant x and x + ^x from the left end of the beam (Fig. 
187). We have seen (Art. 93) that Q is equal to the algebraic 
sum of the vertical forces to the left of x, likewise Q + AQ is 
the sum of those to the left of x + Aa;. Then \Q must be merely 
the additional force exerted between x and x-\-^x. If there is 
no concentrated load in this interval, this force will be l^x 
multiplied by L, the average value of L in this interval. 

We have, then, ^.Q/^x — L; or, since this will approach L as 
Ax approaches zero, 



dQ 
dx 



L, 



(1) 



174 Elementary Mechanics 

where L is the rate of loading, i. e., the load per unit of length 
at the point desired. This of course does not apply at a point 
where there is a concentrated load; but the same discussion 
covers this case also. For if in the interval Ax there is a con- 
centrated load W, AQ will then he W-\-LAx; that is, in addition 
to the continnons change in Q due to the distributed load of 
intensity L there is a sudden change of magnitude W. If we 
write equation (1) in the integral form, 



Jo 



L dx. 



we can include both kinds of loading if we understand the 
integral in a broad sense to include the addition of concentrated 
loads. The rate of change in the shearing force is equal to the 
continuous load per unit of length and any concentrated load 
produces a sudden change in the shearing force of an equal 
magnitude and direction. 

To obtain the relation betAveen M and Q, consider the resul- 
tant of all vertical forces which act to the left of a point distant 
X from the left end. This (by Art. 93) is equal in magnitude 
and direction to the shearing force Q at this point. Let its point 
of application be at a distance x from the left end. Then its 
moment about x is 

M={x-x)Q 

and this is the bending moment at the point x. If we denote 
by M-\-AM the bending moment at a new point x + Ax, this will 
consist of two parts, the original force acting at x and the 
additional force AQ acting in the interval Ax. The arm of the 
original force will now be x-\-Ax — x and that of the new force 
AQ will be some fraction of Ax, BAx, (O^^^l). 

The new moment will then be 

Jf + Ail/= {x^Ax-Zc) Q + BAx . A^. 



Deformable Bodies 175 

Hence 
and 

For a continuous loading AQ approaches zero with Ao: and hence 
dM _^ 

or 

M=\Qdx. 

100. As in the case of the shearing force, the integral proper 
takes account only of the case where the load is continuous. It 
is usual to apply this integral separately to each interval in 
which there is no discontinuity. Its use and that of the integral 
expression for the shear are best explained by an example. 

Example 1. — A simple beam 20 feet long carries a variable 
load on the left half wbose intensity is proportional to the square 
of the distance from the left end and varies from zero at the end 
to 300 pounds per foot at the middle. On the right half it 
carries a uniform load of 300 pounds per foot and in addition a 
concentrated load of 4000 pounds at a point 2 feet to the right 
of the middle. Find the general expressions for the shearing 
force and bending moment at all points of the beam. 

Solution. — The intensity of loading is of the form A\t- on the 
left half. Since this equals 300 when a; =10, we have A; = 3. If 
R^ and Ro are the supporting forces, 

fio 

R^^R^_^\ 3a;- d?; + 10x300 + 4000 = 8000. 
Taking moments about the left end, 

20i?o= T' ?>x^ dx+lQ> X 300 X 15 + 4000 x 12 



Jo 



= 7500 + 45000 + 48000 = 100500. 
i?. = 5025. 



176 Elementary Mechanics 

Hgiicg 

i?,rr:8000-5025 = 2975. 

In the first interval, Oia;ilO, we have L=— 3a:-, neojative be- 
cause the loading is a downward force. 

Now when .t = 0, Q = R^ = 2dlb, since R-^ is then the only force 
on the left. Hence 

(7=2975, 
and 

g = 2975-.T^ (1) 

Integrating again, 

M = 2^1^x~lx^ + C^. 
Since the beam is freely supported, M — O at the ends. Hence 
Ci = and 

i¥ = 2975.T-iT^ (2) 

Before proceeding further let the student note that the equa- 
tions (1) and (2) hold only in the first interval, i.e., when 

0<TglO. 

In the second interval, 10irril2, L=-300. 
.•.(?= -300.T+C.. 
Co must have a value that will make (?(10) have the same value 
as that given by (1), namely 1975. 

.-. g(10)=1975=-3000 + Co. C2 = 4975, 
and 

§ = 4975-300 .T. ' (3) 

Integrating again and determining the constant for M in the 
same way, 

M = 4975.^•-150.^'--7500. • (4) 

In the third interval, 12^.Ti20, we have the same distributed 
load, but a change in the equations because of passing over the 
concentrated load of 4000. This causes a sudden drop in Q, so 
that in this interval 

gr=975-300a;. (5) 

Integrating this and determining the constant so that il/(12) 
will agree with equation (4), 

il/ = 975a;-150a;" + 40500. (6) 



Deformable Bodies 



177 



We may check the correctness of the work by observing in (5) 
and (6) that Q(20) = -5025= -B. and J/ (20) =0. 

The shearing force is given by equations (1), (3), and (5), 
each applying in its own interval, and the bending moment 
similarly by equations (2), (4), and (6). 

Example 2. — Draw the S. F. and B. M. diagrams, for the 
beam in the preceding example. 

Solution. — The cun-e for Q in the first interval, 0^2%10, 
equation (1), starts with ^ = 2975 and the slope zero, and ends 
with () = 1975 and the slope —300, where a:=10. Since x is 
measured in feet, Q in pounds, and M in pound-feet, there is no 
connection between the scales by which they are represented. So 




Fig. 1S8. 



each may be chosen at convenience independently of the others. 
When it is said that the slope of the Q curve is —300 when 
a;=:10, this merely means that the curve is descending at the 
rate of 300 units on the pound scale for each unit advance on the 
foot scale. In each of the other two intervals Q is of the first 
degree and so the graph is a straight line, which is easily drawn 
as soon as we have two points on it (Fig. 188). The graph of 
M may be drawn directly from the equations (2), (4), and (6), 
but we should note that its slope should always be equal to the 
height of the Q curve for the same value of x. From this we 
can see that no part of the graph of J/ is a straight line ; that 3/ 
increases from x = to .t=12 while Q is positive, whereas it 
decreases after that because 



is negative. This at once tells us 



12 



178 



Elementaey Mechan-ics 



that the greatest bending moment occurs when x—12, and is 
equal to 30,600 ponnd-feet (Fig. 189). Note that at this point 
the graph of Q crosses the axis. 




101. Dangerous Section. — In designing a beam it is necessary 
to make it strong enough to resist the greatest bending moment 
that may be applied to it, and this of course means the greatest 
numerically, regardless of sign, since the sign tells us the direc- 
tion of bending. The point where this numerically greatest 
bending moment occurs is called the dangerous section. Thus, 
in the example solved just above, the dangerous section occurs 
where a;=12 feet, since there the value of ilf (30,600 pound-feet) 
was found to be greater than at any other^ section of the beam. 
In finding the dangerous section, it is important for the student 
to realize the necessity of having a clear idea of the general 
character of the bending-moment curve rather than applying any 
mechanical rule. 

Example 1. — A cantilever beam 6 feet long bears a concentrated 
load of 1500 pounds at its outer end. Draw S. F. and B. M. 
diagrams, and determine the dangerous section and greatest 
bending moment. 

Solution. — Let us take the origin at the wall end. The sup- 
porting force must be equal to the load 1500, and of course acts 
upward; hence the S. F. is initially 1500, and.since there is no 



Deformable Bodies 



179 



distributed load it is constant^ and the graph is a horizontal 
line, § = 1500. Integrating this, M=1500x + O; and as there 
can be no bending moment at the free end, ilf = when x=6. 
Hence (7= -9000, and If = 1500x-9000 (Fig. 190). 




Pig. 190. 



It is easy to see that the bending moment at the wall due to 
a load of 1500 pounds six feet out should be numerically 9000 
pound-feet. Actually the sign agrees with the convention of 
Art. 97 ; but it is usually a little safer to determine the constant, 
in the way we have done. The dangerous section is here obvi- 



eSOOIbs. 




^V-SOO\hrH 



b 
-300lbs. 



Fig. 191. 



ously at the wall, and the beam must be strong enough to with- 
stand a B. M. of 9000 pound-feet. 

Example 2. — Consider a cantilever beam 8 feet long, carry- 
ing a uniformly distributed load of 100 pounds per foot run. 
A vertical cable, attached to the outer or free end, is tightened 
until it sustains a load of 300 pounds (see Fig. 191). To 



180 Elementary Mechanics 

find the daiigorous section^ note iliat since 300 ])oiiiu]s of the 
total load is carried at B, 500 poiiiuls is carriml at the wall, 0. 
Hence (^ = 500— IOO.t, giving- eh for the S. K. curve. Then 
il/ = 500a;-50:i:2 + (;; and since l/ = for a; = 8, we find C = 
-800. Tluis ilf = 500a; -50.7;- -800. Now the highest point on 
the bending-nionient curve occurs at a, where C^ = 0; tliat is, 
where x—b feet. This gives yla= +450 pound-feet, which is 
the greatest positive value of M. But at the wall we have M = 
— 800 pound-feet, which is nvmerically greater than the positive 
value Aa; hence the dangerous section is at the wall. 

Problems 

320. Find ili(> Tnnxinuim !*>. M. in Prob. 317. 

321. Two l()('()in()li\(> \vlic(>ls, 5.] feet apart and weighing 
25,000 pounds each, roll over a sini[)ly supported beam, span 30 
feet. Find position of wluu'ls for greatest B. M. and the position 
for the greatest reaction. 

322. Find the niaxiniuni W. !\1. in Prob. 311). 

102. Distribution of Stress. — We have hillicrto considered 
only ilic iolal vWvvl of the strc^ss ncross any vertical section 
of a l()ad(>(l beam without inquiring how the stress is dis- 
tributed, in tlic case of shearing stress we shall not make that 
inquiry, for the reason that it is only in beams that ai"e relatively 
very short that failure* by shear is a possible cause of rupture. 
For bca!ns (•!' all ordinary shapes, failnrc* will occur by bending 
long Ixd'oi'c any dnn^crous shearing stress is j)ro(luced. Failure 
by bending is due to an excc^ssive tensile or compressive stress 
in some part of the scH'tion. To determine the distribution of 
this stress we must take acconnt of tlu* nature of the strain pro- 
duced by the loading; that is, the actual distortion of the beam 
from its original foi'm. ^Fhis distortion in a properly designed 
beam is rebitively vcM'y small; but its nature and magnitude 
must be considered for the purpose, anu)ng others, of determin- 
ing the distribution of the normal stress. 



Deformable Bodies 



181 



103. Bending. — When any part of a beam is acted on by a 
bending moment but no vertical shearing force, it is said to be 
subject to simple bending, e. g., the middle section of the beam 
in Fig. 192. 

In this case we assume (Bernoulli's assumption) that when 
the beam is bent, transverse sections originally plane remain 
plane, and also remain perpendicular to the longitudinal fibers. 
The word " fiber '' here refers to a prism of infinitesimal cross 
section originally parallel to the axis of the beam. There is no 
implication that the material of the beam is of a fibrous nature. 



£ 



W 



V 



W 



\/ 



i 



Fig. 192. 



It follows from this assumption that the fibers on the concave 
side are shortened and so subjected to a compressive stress and 
those on the convex side lengthened and hence under tensile 
stress. Somewhere between these there must be a surface in 
which the fibers are neither elongated nor compressed. This is 
called the neutral surface. A vertical plane through the original 
position of the axis of the beam is called the ylane of lending. 
The intersection of this with the neutral surface is called the 
elastic curve. Any transverse section meets the neutral surface 
in a line which is called the neutral axis of that section. If, as is 
commonly the case, the beam is of uniform cross section through- 
out its length, the neutral surface will be plane before bending, 
cylindrical after, and always perpendicular to the plane of bend- 



182 



Elementary Mechanics 




Fig. 193. 



Whatever the shape of the elastic curve as a whole, it will have 

at any point a center of a curvature, 
and any short arc of it will be very 
closely approximated by a circle with 
this center (Fig. 193). Any two 
transverse sections cut the plane of 
bending in radii of this circle. Let 
its radius be R. Then any arc of 
the circle which subtends an angle 6 
is of length R6. Any other longi- 
tudinal fiber between the same two 
transverse sections has the same 
original length. If it is at a distance y from the neutral surface 
its strained length will be {R±y)0 according as it is toward 
the convex or concave side of the neutral surface. In either 
case its change of length is yO and this divided by RO the original 
length gives y/R as the strain. Hence by Hooke's Law, the 
stress in the fiber is p — Ey/R. Thus the stress in any fiber is 
proportional to its distance from the neutral surface. 

104. Location of the Neutral Axis. — We have already seen 
(Art. 96) that the total horizontal component of the normal 
forces across any section is zero. But since the differential ele- 
ment of this force is p • dA, the total 
force is j p • dA or \ E/R -y -dA 
over the section. Note that y wdll 
be positive on one side of the neutral 
axis and negative on the other, and 
that corresponding to this change of 
sign the normal force acts in oppo- 
site directions on the two sides of 
the neutral axis (see Fig. 194). 




Fig. 194. 



Deformable Bodies 183 

Since E and R are constant for the integration, we have 

where y is the distance from the neutral surface to the center of 
area of the section. Hence y — ^, or the neutral axis of any 
section passes through its center of area. 

105. Moment of the Internal Couple. — The total moment 
about the neutral axis of the normal force across the section will 
be 

Sy.p.dA = ^\y^~dA = ^^.L 

where I is the so-called " moment of inertia " of the area of the 
section about its neutral axis. Since any portion of the beam 
is in equilibrium, the moment of the couple formed by the in- 
ternal stress across any section must be balanced by the moment 
of the external forces on one side of the section. This last we 
have studied in the preceding articles and called M. Hence we 
have M — EI/R. Combining this with the result of Art. 103, we 
have 

V-M_E 

y~T-R ^^^ 

These are the most important and fundamental formulae in the 
theory of bending. The student should make quite sure before 
proceeding that he has clearly in mind the exact meaning of each 
of the six letters involved. The value of the second fraction is 
known at any cross section if we know the loading of the beam 
and the dimensions of the cross section. Equating this to the 
first fraction enables us to determine the stress at any distance 
from the neutral axis, in particular at the extreme fiber, i. e., 
the farthest from the neutral axis. The relation M/I — E/R 
enables us when we know the material of the beam (which de- 



184 Elementary Mechanics 

termines J5') to calculate the degree to which the beam is dis- 
torted/ or its curvature. 

106. Units of Measurement. — In using the equation of the 
preceding article care must be taken to be consistent in the use 
of units. We commonly measure 'p in pounds per square inch; 
therefore y should be in inches. The quantity / used here must 
be carefully distinguished from what is properly called the 
moment of inertia. The latter involves density or mass, and has 
the dimensions ML"^. The / used here is \ifclA and consequently 
has the dimensions L^, being quite independent of density and 
depending only on the dimensions of the section. As the latter 
are regularly given in inches, I will be in inches. To be con- 
sistent with this, then, M must be in pound-inches. 

Example 1. — What will be the greatest normal stress in a can- 
tilever beam 8 feet long, 6 inches deep, and 2 inches wide pro- 
duced by a weight of 500 pounds at its outer end ? 

Solution. — Here the maximum M is at the wall section and is 

8 . 500 = 4000 Ib.-ft., or 48000 Ib.-in. 

The neutral axis of a section is its horizontal diameter. Hence 
for a rectangle about a diameter 

7=3-126/^3^3?^. 2. 216 = 36 in.^ 
Hence 

p/3 = 48000/36, or p = 4000 lbs. per sq. in. 

Example 2. — A simple beam 8 feet long and 4 inches wide is 
to support a uniform load of 600 pounds per foot run. W^hat 
must its depth if the stress is not to exceed 2000 pounds per 
square inch ? 

Solution. — The total load is 4800, and hence each supporting 
force is 2400. 

g = 2400-600.T, 
If = 2400a; -300x^ 



Deformable Bodies 185 

When x = 4:, If = 4800 (the maximum value) in pound-feet or 
57,600 in pound-inches, 

7=3-V(W)=Fl^ 

y = h/2 and p=2000. 
Therefore, 

2000/(V2) =57600/(p3) ; 

h = V{S> 57600)7(2 . 2000) = V43.2 = 6.57 in. 

ProMems 

323. How much stronger is a 2" X 10" joist placed on edge 
than a square beam of the same material and equal cross section ? 

324. Find the dimensions of the strongest beam that can be 
sawed from a log of diameter d. Compare the strength of this 
beam with that of the original log. 

325. A wooden cantilever beam is 2 inches wide, 5 inches deep 
and 8 feet long. How great a weight can it support at its free 
end, allowing a factor of a factor of safety of 8 

326. A steel engine shaft of constant cross section rests on 
bearings 6 feet apart and supports a 10-ton flywheel midway 
between the bearings. Find the diameter of the shaft, allowing 
a factor of safety of 10. 

327. The interior and exterior radii of a hollow cylindrical 
shaft are r^ and n respectively. Compare the strength of this 
shaft with that of a solid shaft of the same material and equal 
cross section. Compare the strength of it with that of a solid 
shaft of the same external diameter. 

328. A simple wooden beam 3" wide, 6" deep and 15' span 
has a load of 150 pounds at the middle. Compute the factor of 
safety. 

329. A simple wooden beam 6" wide, 8" deep, span 10', carries 
two equal concentrated loads at 2 feet and 8 feet from the left 
end. Find these loads for a factor of safety of 8. 

330. A rectangular wooden beam 8 feet long, 10 inches deep, 
supported at the ends, carries loads of 3 tons and 2 tons at 2 feet 
and 5 feet respectively from the left end. If the maximum stress 
allowed is J ton per square inch, find the width of the beam. 

331. A steel beam 12 feet long, 6" wide, and 1" deep is fixed 
at the left end and unsupported at the right and weighs 20 pounds 



186 Elementary Mechanics 

per foot run. What weight at the free end will destroy the beam, 
if the nltimate stress is 100,000 pounds per square inch? 

M I 
107. Section Modulus. — Equation (5) may be written — — ~ 

If y has the value, usually called c, for the extreme fiber, we 
get the ratio I/c called the section modulus. This is a quan- 
tity depending entirely on the shape and size of the cross section, 
which gives at once the ratio of the bending moment at any sec- 
tion to the maximum stress in that section. It can be found 
worked out in handbooks for a number of standard sizes and 
shapes of beams. In a practical problem of design, M and p 
are known. The section modulus is obtained from the relation 
l/c — M If. It is then merely a question of picking out of the 
handbook the proper size for the required section modulus. 

Example. — A simply supported steel 7-beam, span 20 feet, 
carries a uniform load of 500 pounds per foot run. What should 
be its depth ? 

Solution.— The greatest i¥ = 500 • 20 • 20 . 12/8 pound-inches. 
If a factor of safety 5 is used p = 13,000 pounds per square 
inch. Then 

I _ 500.20.20.12 _oQno 
c- 8.13,000 -^^•^^- 

Eeference to a steel handbook gives the proper depth as 9 
inches with weight 35 pounds per foot. 

The weight of the beam gives an additional uniform load. To 
check the result the load per foot run must be taken as 535 
pounds. Then 

I ^ 535.20.20.12 ^ 

c 8.13,000 "" ■ • 

Since the section modulus of the 9-inch beam is 24.8 and thus 
larger than required, the design will not have to be changed. 

Problems 
332. What should be the section modulus of an I-beam, 20 
feet long, which supports a uniform load of 450 pounds per foot 
run and two concentrated loads of 1500 and 2700 pounds at 6 
and 14 feet respectively from the left end ? 



Deformable Bodies 187 

333. Wooden floor joists of 16-foot span and spaced 14 inches 
from center to center, are required to support a floor load of 100 
pounds per square foot. What should be the size of the joists if 
the allowable stress is not to exceed 1100 pounds per square inch? 

331. A balcony is to project 5 feet from a wall and be sup- 
ported by wooden beams spaced 2 feet from center to center. If 
the floor load is to be 250 pounds per square foot and the fiber 
stress must not exceed 900 pounds per square inch, find the most 
suitable size for the beams. 

335. An IS" I-beam, weighing 90 pounds per foot, carries a 
uniform load of 50 tons. Compute its factor of safety if the 
span is 6 feet; also if the span is 9 feet. 

336. A steel channel, span 20 feet, flanges horizontal, carries a 
uniform load of 13,500 pounds. "What should be its size? 

337. What size I-beam should be used to support the loads in 
Prob. 321? 

108. Superposition. — If a beam is subjected to any particular 
loading, and the bending moment, or stress, etc., worked out at 
any point, and then the same thing done for another loading, 
it will be found that if both loads, are applied at once we must 
at every stage of the work add together the two results previously 
found. This method is called superposition. It has its most 
important application in the use of a handbook. There a 
number of simple loadings are worked out, and one may solve a 
great many problems by treating them as combinations of these 
cases. 

" Prohlons 

338. If the beam in Prob. 318 weighs 20 pounds per foot 
run and is loaded as stated, draw the B. M. and S. F. diagrams. 

339. A cantilever beam, 8 feet long and weighing 5 pounds 
per foot run, carries a concentrated load of 300 pounds at the 
free end and another concentrated load of 200 pounds at its 
midpoint. Draw the B. M. and S. F. diagrams. 

310. A steel girder, 18 feet long and weighing 10 pounds per 
foot run, is supported at its ends and carries concentrated loads 



188 



Elementary Mechanics 



of 4000, 1000, and 600 pounds at 5, 9, and 13 feet respectively 
from the left end. Draw the B. M. and S. F. diagrams. 

341. A beam 20 feet long, supported at its ends, carries a 
uniform load of 60 pounds per foot run, and also 800 pounds at 
6 feet from left end and 500 pounds at 16 feet from left end. 
Draw the S. F. and B. M. curves and give the values of S. F. 
and B. M. at 14 feet from left end. Find the greatest B. M. 

109. Torsion. — If a cylindrical bar is held fast at one end and 
a couple applied to it at the other, tending to twist the bar about 
its axis, the bar is said to be subject to torsion. The twisting 
moment tends to rotate adjacent sections of the bar in relatively 
opposite directions about the axis. Thus the stress acts tan- 




FiG. 195. 



gentially on the section of the bar, and, therefore, torsion is a 
kind of shear. We shall confine our attention to bars with cir- 
cular cross sections. 

Suppose a force F with arm a tends to turn a shaft of radius r 
about its axis as shown in Fig. 195. Take a point B on the 
shaft and suppose a perpendicular section to be passed through 
it. To keep the part of the shaft to the right of the section in 
equilibrium, the action of the external forces must be resisted 



Deformable Bodies 189 

by the internal stresses acting at the section. Hence, taking 
moments about the axis, the sum of the moments of the external 
forces acting on the shaft is eqnal to the sum of the moments of 
the internal forces. The former, in this case F - a, is called the 
torque and denoted by T ; the latter is called the resisting 
moment. Thus we say 

Torque =:Eesisting moment. 

If in Fig. 195 an element of the cylinder AB is twisted into 
the curve AB' , we will denote the angle BAB' by </> and the angle 
BOB' by 6, The angle </> is called the shearing strain; 6 is called 
the angle of twist. We find by experiment that q = G4>, where G 
is a constant and is called the '' shearing modulus of elasticity " or 
'"'modulus of torsion.^^ From the figure we see that l(f> = r9. 

Thus r=-^^, and we see that the stress is proportional to the 

distance from the axis and reaches its maximum value at the 
surface. Thus if q' is the stress at a distance p from the center, 
q/r—q'/p and q' = pq/7\ On a circular ring of radius p and width 
dp the total stress is q'27rpdp. Taking moments about and 
integrating over the whole area, we find the resisting moment. 
Thus 



T^i^yq'p^-dp=^-^ . \^ydp = 



171^ q 



2r 

But 7rrV2 is the '' Polar Moment of Inertia " of the area of the 
section about and we may write 

which is analogous to the formula for Bending Moment 3/ = 

v/y ' I- 

Substituting the value of q in terms of 6 and solving for 
we have 



Therefore 



190 Elemextaey Mechaxics 

which shows that the angle of twist is proportional to the twist- 
ing moment. 

Example. — What twisting moment in ton-inches can a steel 
shaft 3 inches in diameter transmit safely? Use a factor of 
safety 10. 

Solution. — From the table 

(/ = 50000 lbs. per sq. in. 
Ip = 7T{^)^ 'i in inch units. 

r^M^MOOlMfV ^,6,500 Ib.-m. 

= 11.8 ton-inches. 

Prohlems 

312. Compare the torsional strength of a hollow shaft of radii 
7\ and r., with that of a solid shaft of the same cross-sectional 
area. 

313. If 7\ = |-?-o compare the strengths of the two shafts in 
the above problem. 

311. What torqne wonld be required to twist off a steel spindle 
1 inch in diameter? 

315. Compnte the stress in a hollow shaft, ontside diameter 
18", inside diameter 10". subjected to a torqne of 250 ton-feet. 

316. A solid ronnd shaft is subjected to a torqne of 300 ton- 
feet. Find the diameter if the stress is limited to 6000 ponnds 
per square inch. 

317. A steel wire 0.18" in diameter and 20" long is twisted 
through an angle of IS -5 by a moment of 20 pound-inches. 
Find the shearing modulus of elasticity. 

Feriew Prohlems 

318. A stone pillar has to carry a load of 8 tons and its cross 
section is to be a rectangle whose sides are in the ratio 3:1. If 
the stress allowed is 120 pounds per square inch, find the dimen- 
sions of the smallest section which will suffice. 

349. A steel rod 2" in diameter and 8 feet long stretches j^" 
when subjected to a pull of 11 tons. Find the modulus of elas- 
ticity. 



Deform ABLE Bodies 



191 



350. A steel tube (inner diameter 8 inches, outer diameter 9 
inches) is 5 feet long and is subjected to a pull which lengthens 
it YQ inch. Find the pull in tons. 

351. A hollow steel column is 8' long, 10" square outside, and 
square in section within. Under a tensile force of 700 tons it 
stretches yV'. Find the thickness of the walls of the column. 

352. What is the length of an iron rod (vertical) which will 
just carry its own weight? {p = 7000 pounds per square inch.) 

353. The steel plates of a girder are 1" thick and are double 
riveted with double butt straps. Eivets are f " in diameter. If 
the shearing stress is f the tensile stress, find the pitch for equal 
strength. Also find the efficiency of the joint. 

354. A copper pipe, 2" internal diameter, 2 J" external diam- 
eter, is subjected to an internal pressure of 1200 pounds per 
square inch. Find the stress in tons per square inch. 

355. A beam 4" wide and 8" deep is subjected to a bending 
moment of 160 ton-inches; what is the maximum fiber stress? 

356. A steel rod, 10 feet long, 3 inches in diameter, is sup- 
ported at its ends ; what is the greatest central load it can carry, 
if stress is limited to 4 tons per square inch ? 

357. A T-beam, 12 feet long 

and with section shown in Fig. ^>, 

196, carries a central load of 24 

tons ; neglecting the weight of the 

beam, find the maximum fiber 

stress. 

358. What is the radius of the 
smallest circle into which an iron 
rod 1^" in diameter can be bent 
without injury, the stress being 
limited to 3| tons per square inch ? 

359. A log is in the form of a 
triangular prism, its cross section 
being an equilateral triangle, 30" 
on a side. Find the breadth and 
depth of the rectangular beam of 
greatest strength which can be 
sawed from this log. 



3" 

3" 


2' 
2 


3' 
3' 




6' 



Fig. 196. 



192 Elementary Mechanics 

360. A steel I-beam is 24 feet long, and carries 28000 pounds 
at its center. Its flanges are 7" by f and the web is 24" by -J". 
If the weight of the beam is neglected find the maximum fiber 
stress. 

361. A plank 15 feet long, supported at its ends, carries 400 
pounds 4 feet from the left end, 300 pounds 8 feet from the left 
end, and 250 pounds 10 feet from the left end. Draw the S. F. 
curve and find the S. F. 7 feet from the left end. 

362. A beam 16 feet long, supported at its ends, carries a 
uniform load of 80 pounds per foot run. Draw the S. F. curve 
and give the vahie of the S. F. at 12 feet from the left end. 

363. A beam 10 feet long has its left end fixed in a wall and 
supports a load of 1200 pounds at its free end. Draw the S. F. 
curve and find the value of the S. F. at 10 feet from the left end. 

364. A beam 16 feet long, supported at its ends, carries a load 
of 60 pounds per foot run, and has concentrated loads of 1 ton 
and 3 tons at 4 feet and 10 feet, respectively, from the left end. 
Draw the S. F. curve, evaluate the S. F. at 6 feet from the left 
end and find where the S. F. = —5000 pounds. 

365. An oak beam 12 feet long, 18 inches wide and 12 inches 
deep, floats in sea water. It is loaded at the center with a weight 
which just immerses it wholly. What is the greatest value of 
the shearing force and where does it occur? 

366. A beam 12 feet long has its left end fixed in a wall, and 
it carries a uniform load of 80 pounds per foot run, together 
with a concentrated load of 800 pounds 7 feet from the wall. 
Draw the S. F. curve and evaluate the S. F.at 8.5 feet from the 
wall. 

367. A beam 12 feet long, supported at its ends, carries a load 
of 1600 pounds 9 feet from the left end. Draw the S. F. and 
B. M. curves, give the value of B. M. at the center and the great- 
est value of B. M. 

368. A beam 16 feet long, supported at its ends, carries a 
uniform load of 80 pounds per foot run. Draw the S. F. and 
B. M. curves, and give B. M. at the center and 14 feet from left 
end. 

369. A beam 12 feet long, fixed at left end and unsupported 
at the right, carries a uniform load of 100 pounds per foot run. 
Draw the S. F. and B. M. curves and give the maximum value 
of B, M. and its value 8 feet from left end. 



Deform ABLE Bodies 193 

370. A steel bar, 4 feet long, 1 inch wide, is supported at its 
ends. It has a weight of 200 pounds at its center and one of 
600 pounds at 3 feet from the left end. Neglecting the weight 
of the bar, find its depth if stress is limited to 12,000 pounds per 
square inch. 

371. A steel beam is 12 feet long, 8 inches square, supported 
at its ends, and carries the following loads : 24 tons, 4 tons, and 
20 tons at 5 feet, 7 feet, and 10 feet, respectively, from the left 
end. Neglecting the weight, find the maximum stress due to 
bending. 

372. A beam, 13 feet long, supported at the ends, carries loads 
of 1000 pounds, 800 pounds, and 1140 pounds at 3 feet, 8 feet, 
and 10 feet, respectively, from the left end. Draw the S. F. and 
B. M. curves and find the B. M. 9 feet from the left end. 

373. A beam 20 feet long, supported at the ends, carries a 
load which uniformly increases from 20 pounds per foot run at 
the left end to 60 pounds per foot run at the right end. Draw 
the S. F. curve, find the maximum value of the S. F. and the 
point where the S. F. = 0. 

374. A wooden beam of rectangular section is 8 inches wide; 
find its depth, if the maximum B. M. is 18 ton-feet and the stress 
allowed is f ton per square inch. 

375. What i^ the diameter of the smallest circle with which 
^-inch steel wire may safely be coiled, keeping the stress within 
5 tons per square inch ? 



13 



194 Elementary Mechanics 



CHAPTER VIII 

RECTILINEAR MOTION 

110. Velocity. — Previous chapters have dealt mainly with 
bodies at rest, any mention of motion being incidental. This 
and the succeeding chapters will deal with motion and moving 
bodies, beginning in this chapter with motion in a straight line.* 
The average velocity of a body moving in a straight line during 
a given time interval is defined as the distance moved divided 
by the time elapsed. If this quotient is the same, regardless of 
the choice of the interval, the velocity is said to be constant. If 
this is not the case, we define the velocity at any instant to be the 
limiting value of the average velocity for an interval beginning 
with that instant, as the length of this interval decreases and 
approaches zero. If s is the whole distance traversed in the time t. 
then the average velocity is the increment of distance As divided 
by the increment of time A^^ and the instantaneous velocity is 

lim — or -^ , the derivative of s with respect to t. 
At=oAt cW -^ 

The numerical measure of the velocity depends, of course, 
upon the units of distance and time employed, and the unit of 
velocity must name both of these, e. g., miles per hour, or feet 
per second. If the unit of length is changed, the numerical 
measure of the velocity changes in the same way as the measure 
of the length : that is, since the number of feet is multiplied by 
12 to get the number of inches, the number representing the 
velocity in inches per minute is 12 times that in feet per minute. 
A change in the unit of time, however, produces the opposite 
effect. Thus in changing minutes to seconds we multiply by 60 ; 
but in changing feet per minute to feet per second, we must 
divide by 60. We say then that velocity is of dimensions 1 



Eectilinear Motion 195 

in length and —1 in time^ L^T'^, and the units are commonly 
written in the corresjDonding form, mi/hr, ft/sec, etc. 

Problems 

376. How many feet per second is a velocity of 15 miles per 
hour? 

377. A piece of wood floating on a river takes 5 seconds to 
pass under a bridge 14 yards wide. Find the speed of the stream 
in (a) feet per second, (b) miles per hour. 

378. It takes a stone 3 seconds to fall 151 feet. What is its 
average velocity in miles per hour during that time? 

379. Express a velocity of 1094 yards per hour in centimeters 
per minute. 

380. Express a velocity of 7^ feet per second in kilometers 
per hour, given 1 meter = 3.281 feet. 

381. If telegraph poles by the side of a railway are a yards 
apart and n poles are passed in t minutes, find the velocity of the 
train in miles per hour. 

382. If the distance in feet passed over by a moving body in t 
seconds is given by the equation 

s = 3^3^4t + 2, 

find the velocity at the end of 3 seconds, 2 seconds, -J second. 

383. If the distance is given by the equation 

5=10 cos I t, 

find the velocity when i5 = 0, }^, 1, If, 2, 2^, 3, 3|, 4. 

111. Acceleration. — If the velocity i) is a variable, the change 
in velocity Au in a time interval Af, divided by this M, is the 
average acceleration. The limit of this as A^ approaches zero 
is the acceleration at the instant at which A;^ begins, and is 
therefore equal to dv/clt or d^s/dt^. It is regularly denoted by 
the letter a. The unit of acceleration must be a unit of velocity 
(or change in velocity) per unit of time, and hence the unit of 
time occurs twice, both times in the denominator. The time 
units need not be the same in the two cases, but frequently are. 



196 Elementary Mechanics 

Thus if a train at one instant is moving with a speed of 30 
mi/hr and 5 minutes later with a speed of 45 mi/hr, its change 
of velocity is 15 mi/hr, but, as this has occurred in 5 minutes, 
the average change per minute is 3 miles per hour; that is, the 
average acceleration is 3 miles per hour per minute. If the 
length unit or either of the two time units is changed the measure 
of the acceleration is changed accordingly. Thus 3 miles per 
hour per minute is 15,840 feet per hour per minute, or 2G4 feet 
per minute per minute, or 4.4 feet per minute per second, or 
.073 feet per second per second. Note that the measure of the 
acceleration in feet per minute per minute must be divided by the 
square of 60 to obtain the acceleration in feet per second per 
second. The dimensions of acceleration, therefore, are L^T~^ and 
the units may be written ft/sec^, cm/sec", etc. 

Prohlems 

384. Express an acceleration of 800,000 feet per hour per hour 
in miles per minute per minute. 

385. An acceleration of 27,000 miles per hour per hour is 
how many feet per second per second? 

386. If a point acquires in 15 seconds a velocity of 30 miles 
per hour, what is the acceleration in feet and seconds ? 

387. A body has a velocity of 33 miles per hour; 3 minutes 
later its velocity is 60 miles per hour. Express the average 
acceleration in feet and seconds, 

388. If s, the distance traversed by a moving body in time t, 
is given by the equation 

s = at + W + ci\ 

find velocity and acceleration at time t. 

389. When s and t are connected by the equation 

s^aco^ {h-\-lct), 

find the velocity and acceleration at time t. 

390. If a point moves so that s=Vt, show that the acceleration 
is negative and proportional to the cube of the velocity. 



Eectilixear Motion 197 

391. If a point moves so that in t seconds s — lO log j—j feet, 

find tlie velocity and acceleration at the end of 1 second. x\t the 
end of 16 seconds. 

393. The motion of a particle in a straight line is expressed by 
the equation s = 5 — 2 cos'-f. Express the velocity and acceleration 
at any point in terms of s. 

112. A very important form for the acceleration in terms of 
derivatives is obtained by the elimination of /, thus 
_ dv _ dv ds _ d£ 
~ dt ~ ds dt ds 
Hence 

where Vo is the velocity when 5 = 0, or 

i'-z=Z'o- + 2 a-ds. 
Jo 

If a is a constant, this becomes 

1'-= Vo--\-2aS. 

It must be borne in mind, however, that this last expression 
applies only when the acceleration is constant. 

Problems 

393. A body starts with a velocity of rt feet per second andt 
moves with an acceleration of 1 foot per second per second. In 
what time will it acquire a velocity of 30 miles per hour? 

39^. A body starting from rest, and moving with uniform 
acceleration, describes ITl feet in the tenth second. Find its 
acceleration. 

395. A point starts with a velocity of 100 centimeters per 
second and moves under a retardation of 2 T^entimeters per second 
per second. AYhen will its velocity by zero, and how far will it 
have gone? 

396. A point moving with uniform acceleration describes in 
the last second of its motion 9/25 of the whole distance. If it 
started from rest, how long was it in motion and through what 
distance did it move if it described 6 inches in the first second? 



198 Elementary Mechanics 

397. A point moving with uniform acceleration describes 25 
feet in the half-second which elapses after the first second of its 
motion, and 198 feet in the eleventh second of its motion. Find 
the acceleration and initial velocity. 

398. A point starts from rest and moves with a uniform accel- 
eration of 18 feet per second per second. Find the time taken by 
it to traverse the first, second, and third foot respectively. 

113. Relative Velocity. — In determining the position, velocity, 
or acceleration of a body, we must always measure these with 
respect to some body, some system of coordinates or the like, which 
for the time being we regard as fixed. Thus if a man walks 
from one end to the other of a car 40 feet long in 10 seconds, we 
should say that he walks at the rate of 4 feet per second. This 
is his speed with reference to the car. If, however, the car itself 
is moving, his speed with reference to the track is quite different. 
It is easy to see that the velocity of the man with respect to the 
track is the algebraic sum of the velocity of the car with respect 
to the track and that of the man with respect to the car. On the 
other hand, if the velocities of both are given with respect to the 
track, the velocity of the man relative to the car is the algebraic 
difference of their velocities, that of the car (i. e., the new refer- 
ence body) being subtracted. We can formulate the principle 
illustrated by the first case in the equation, velocity of A relative 
to C = velocity of A relative to B + velocity of B relative to 
C, proper attention being given to the signs of the velocities. If 
we solve for the first term of the second member, we get the 
theorem illustrated by the second case, velocity of A relative to 
B=: velocity of A relative to C — velocity of B relative to C. 
The same principle applies to relative accelerations. 

Problems 

399. A train is moving with a speed of 30 miles per hour, and 
another train on a parallel track is going in the opposite direc- 
tion with a speed of 20 miles per hour. What is the velocity of 
the second train as observed by a j)assenger in the first? 



Eectilinear Motion 199 

400. A man can throw a certain body with a velocity of 20 feet 
per second. If he stands on the rear platform of a train moving 
30 miles per hour and throws the body straight backward, what 
velocity will it have with respect to the track when it leaves his 
hand ? 

114. Mass and Momentum,— So far in this chapter we have 
spoken only of motion without reference to the cause of the 
motion or the character of the moving body. We shall now intro- 
duce the idea of mass. Like many other elementary concepts, it 
is not easy to define satisfactorily in a few words. A very rough 
notion of it is obtained by describing it as quantity of matter; 
but this suggests no definite method for its measurement. The 
most fundamental fact about it is this: If a body is defined as 
a definite portion of matter, then the mass of a body remains 
unchanged if no material is added to the body or taken from it. 
This at once distinguishes mass from volume, which changes 
with alterations of pressure, temperature, or chemit;al state. It 
also distinguishes it, but less obviously, from weight, which is 
the measure of the earth's apparent attraction. For it is found 
that the weight of a given body as measured by a spring balance 
varies quite appreciably with changes of latitude and elevation 
above sea-level. This distinction must be made even if we con- 
fine our attention to bodies in the neighborhood of the earth's 
surface; and in astronomy the notions of mass and weight be- 
come entirely distinct, mass remaining fundamental, a "body- 
constant," and weight becoming quite accidental and variable, 
depending on the body's location. A second property of mass is 
that the mass of a body is equal to the sum of the masses of its 
component parts. For practical measurements of mass, we make 
use of a third assumption : that in neighboring localities near the 
surface of the earth, the masses of two bodies are equal if their 
weights are equal ; that is, if they balance each other on the scale 
pans of an ordinary beam balance, it being a matter of experi- 



200 Elementary Mechanics 

ment and observation that two bodies which thus balance each 
other when weighed in one locality will do the same when 
weighed in any other locality. These three assumptions are 
sufficient for us to measure mass, i. e., to express the mass of any 
body as a multiple of some standard mass taken as a unit. There 
are two principal standards of mass : the pound, preserved by the 
British government in London, and the kilogram, preserved by 
the French government, official copies being kept by the Bureau 
of Standards in Washington. It is to be noted that these are 
standards of mass and not of weight. For, although they are 
copied by being weighed, the copy when transported to a differ- 
ent region of the earth's surface may have an appreciably differ- 
ent weight; but its mass is assumed to be still equal to the mass 
of the prototype. The kilogram is 2.2046 pounds. Scientists 
commonly use as a standard mass the gram, or one-thousandth 
part of a kilogram. 

It is possible to express all physical units in terms of three 
fundamental ones. There is some difference in practice as to 
which three shall be used ; but the most usual selection is that of 
length, time, and mass. Every other unit then has " dimensions " 
in the three given units. 

The momentum of a moving body is defined as the product of 
its mass and its velocity. Its dimensions, therefore, are M^L^T~'^. 

115. Newton's Laws of Motion. — The most fundamental rela- 
tions connecting space, time, mass, and force are expressed in 
the traditional form of three laws formulated by Sir Isaac 
Newton as follows: 

I. Every body continues in its state of rest or uniform motion 
in a straight line, except in so far as it may be compelled by 
impressed forces to change that state. 

II. Change of motion is proportional to the impressed force, 
and takes place in the direction of the straight line in which the 
force acts. 



Eectilixear Motiox 201 

III. To every action there is always an equal and contrary 
reaction, or, the mutual actions of any two bodies are always 
ecjual and oppositely directed. 

These laws are generalizations from experience and differ 
from any other physical laws only in being of the widest possible 
application. They are not altogether original with Xewton, 
much of their content having been previously known to Galileo 
and Huyghens. They were originally stated in Latin ; but the 
English translation given above lias become more or less tra- 
ditional. As a consequence, the laws require considerable eluci- 
dation for the modern student. They are included here partly as 
a matter of historic interest and general information. It is im- 
portant, however, that the student should obtam a thorough 
understandino- of their meanino- as stated. 

The First Law is sufficiently clear in its language; but at 
first sight it appears to contradict experience. We commonly 
find that it is necessary to apply a force if we are to keep a body 
moving uniformly in a straiglit line. Actually, however, this 
force is required only to overcome frictional or other resistances, 
as we can readily see if we observe that the force required 
becomes less and less as these resistances are diminished. Thus 
a heavy body sliding on smooth ice, or a large ship moving slowly 
in still water, will continue for some time with only a small 
reduction in speed, and even in these cases there is some resist- 
ance. All bodies in nature are subject to the action of some force ; 
but if a body is acted on by a combination of forces whose re- 
sultant is zero, the First Law is applicable. This law has been 
paraphrased by Kelvin and Tait in the words, '' It is force alone 
A'rhich can produce a change of motion."" It is thus mainly pre- 
paratory to the Second Law, which tells us how force changes 
motion. 

Regarded quantitatively, " change of motion " in the Second 
Law must be understood to mean change of momentum^ not 



202 Elementary Mechanics 

merely change of velocity. Thus equal forces acting on unequal 
masses for a given time will produce unequal changes in their 
velocity, bui equal changes in their momentum. It must be 
observed also that the forces must be assumed to act for the 
same length of time in order that the resulting changes in 
momentum shall be proportional to the forces. 

The Third Law has to do with the nature of a force and states 
that it is something with two ends. That is, if one body is 
pushed upward with a certain force, there is somewhere another 
body which is being pushed downward with an equal force. Or, 
recalling the definition of a force as a push or a pull, we may 
say that a force is that which pushes two bodies apart or pulls 
them together. Thus, if a man's hand pushes against a table 
with a certain force, the table pushes against his hand with 
the same force and in the opposite direction, and this is true re- 
gardless of whether this force is sufficient to move the table or not. 
The equality of action and reaction must not be confused with the 
condition for equilibrium. A body is in equilibrium when it is 
acted on by two equal and opposite forces; but these two forces 
are 7iever action and reaction. For they act on the same body, 
whereas action and reaction always act on different bodies. 
Action and reaction are sometimes spoken of as the two ends of 
a single force, but more commonly as two forces. This is merely 
a question of words, and is unimportant. It is very necessary, 
however, to remember the existence of the reaction, and, if we 
are considering the effect of a force on a system of bodies, to note 
whether the reaction acts on the system or on an outside body. 

116. Conservation of Momentum. — If two bodies act on each 
other for a certain interval of time, we learn from the Third Law 
that the force acting on one is always equal and opposite to the 
force acting on the other, and hence by the Second Law the 
change in the momentum of one is equal and opposite to the 



Eectilixear Motiox 203 

change in the momentnm of the other. If ve confine onr 
attention to motion in one straight line, we may state this in 
the form : whatever momentum the one body gains the other loses, 
due regard being paid to algebraic sign, and the sign of momen- 
tum being the same as that of velocity. The combined momen- 
tum of the two bodies is then the same at the end of the interval 
as at the beginning. If we apply this principle to several bodies 
exerting mutual actions and reactions upon one another, but 
affected by no external force, we see that the action and reaction 
between any two bodies produce ecpial and opposite changes in 
momentum, and hence all the actions and reactions together 
leave the total momentum unchanged. We may then state the 
principle that if a system of bodies is not acted upon by any ex- 
ternal force, its total momentum is a constant. This is known 
as the principle of conservation of momentum. Thus if we know 
that at one instant a mass of 20 pounds is moving east at a 
speed of 5 ft/sec, a mass of 10 pounds is moving east at 3 ft/sec 
and a mass of 2 pounds is moving west at 10 ft/sec, and if later, 
after certain collisions or other mutual actions and reactions, we 
know that the first body is moving east at tt ft/sec and the second 
east at 2 ft/sec we may find the speed v of the third body by 
writing the equation : total original momentum equals total final 
momentum. Thus, taking east as the positive direction, 
20 X 5 + 10x3 + 2 (-10) =20x4-1- 10 x2 + 2r; 
whence v = d, that is, the third body has a final velocity of 
5 ft/sec east. 

Problems 

401. A body of mass 3 pounds moving 13 ft/sec overtakes a 
l)ody of mass 2 pounds moving 3 ft/sec in the same straight line. 
If they coalesce and form one body, find its velocity. 

402. A shot of mass 1 ounce is projected with velocity 1000 
ft/sec from a gun of mass 10 pounds. Find the velocity with 
which the latter beoins to recoil. 



204 Elementary Mechanics 

117. Elastic Impact. — If two more or less elastic bodies col- 
lide, it is found as a matter of experiment that, if the collision is 
not too violent, the velocity with which they recede from each 
other bears a ratio to the velocity of approach which is prac- 
tically independent of the size and shape of the bodies and the 
velocities themselves, and depends only on the materials of which 
the bodies are made. This ratio is called the coefficient of elas- 
ticity, or, better, the coefficient of restitution, for the two sub- 
stances and is commonly denoted by e. Its value varies between 
the limits zero and unity, being zero for perfectly inelastic 
substances such as putty, and nearly unity for two bodies both of 
glass. It will also be zero in the case of one body penetrating the 
substance of the other and remaining imbedded in it, as a bullet 
remaining imbedded in a block of wood. It is to be noted that 
there is no such thing as a coefficient of restitution for a single 
substance unless we understand that to mean the coefficient for 
the impact of tAvo bodies both of that substance. Moreover, if 
we know the coefficient for two bodies of one substance and also 
for two bodies of some other substance, we cannot predict the 
coefficient for two bodies, one being of one of these substances 
and the other of the other. Thus (according to Jeans) the 
coefficient of restitution for iron on iron is about .66, for lead 
on lead, about .20 ; but for iron on lead the coefficient is not even 
intermediate between these values, but is about .14. There is 
here a decided analogy between the coefficient of restitution and 
the coefficient of friction, as the latter is also apt to be smaller 
for unlike substances than for like. 

Since the velocity of approach is the algebraic difference in 
the velocities of the bodies before impact, and the velocity of 
separation the same difference after impact except that the sub- 
traction is performed the opposite way, we may say that the 
relative velocity before impact must be multiplied by —e, to 



Eectilinear Motion 205 

obtain the relative velocity after impact. This fact, together 
with the principle of conservation of momentum, enables us to 
predict completely the motion of two bodies after impact if we 
know their velocities before and their coefficient of restitution. 
Thus if m and m' are the masses of two bodies, u and u' their 
respective velocities before impact, and v and v' their velocities 
after impact, we have from the conservation of momentum 

'mu-{-m'u' = mv-{-m'v\ 

and from the definition of e 

v — v'=—e{u — u'). 

Example 1. — A mass of 20 pounds moving north with a speed 
o'f 8 ft/sec m.eets a mass of 10 pounds moving south with a speed 
of % ft/sec and their coefficient of restitution is -J. What are 
their velocities after impact? 

Solution. — From the conservation of momentum 

20i; + 10i;' = 20x8-10x5. " 

From the coefficient of restitution 

i;-t;'=-i[8-(-5)]. 
Hence, 

v = ^ f/s and v' — S f/s. 

This shows that the first body has its speed much reduced but 
continues in the same direction. The second body, however, has 
its motion reversed, as its original velocity was negative. 

Eaximple 2. — Find the velocities if the two bodies are perfectly 
elastic. 

Solution. — The momentum equation is the same as before, 
but now 6 = 1, and hence 

v-v'=-[8-{-5)]. 
This gives 

7;=-|f/s, i;' = 12J-f/s. 

In this case both bodies have their direction of motion reversed. 



206 Elementaky Mechanics 

Prohlems 

403. A ball A overtakes a ball B moving in the same straight 
line. A has half the mass and seven times the velocity of B. 
The coefficient of restitution between them is J. Find their 
velocities after impact. 

404. A ball of mass 2 pounds impinges directly on a ball of 
mass 1 pound which is at rest. If the velocity of the former 
before impact is equal to the velocity of the latter after impact, 
find the coefficient of restitution. 

405. Two balls have masses 14 and 7 pounds respectively, and 
velocities in opposite directions of 11 and 22 ft/sec respectively. 
The coefficient of restitution is f . Find their velocities after 
impact. 

406. A ball drops from the ceiling of a room and after twice 
rebounding from the floor reaches a height equal to one-half that 
of the room. Find the coefficient of restitution. 

118. The Equation of Motion. — In discussing Newton's Second 
Law it was tacitly assumed that the forces acting remained 
constant during the experiment. We shall generalize the law 
so as to remove this restriction, still confining our attention as 
yet to motion in a straight line. If we are to take account more 
explicitly of the time element, it is necessary to say that the 
change in momentum per unit of time is proportional to the 
force applied, the force remaining constant. Thus if t, v, m, and 
F denote time, velocity, mass, and force, respectively, we shall have 
from the Second Law 

A{mv)/At-XF, 

where A is a numerical constant depending only on the units in 
which t, V, m, and F are measured. If F is variable this equation 
must become 

A<=o A^ at at 

where m passes over the sign of differentiation because it is a 
constant. This result, that mass times acceleration is a constant 
times the force applied, is by far the most important and funda- 



Eectilixear Motion 207 

mental fact in dynamics. The rest of this chapter and mnch 
of what follows will consist of nothing but the development and 
illustration of this equation. The equation in this form, \F = 
ma, is sometimes called the equation of motion, but more cor- 
rectly the differential equation of motion, since a is a second 
derivative, d-s/dt~. The shorter phrase, equation of motion, is 
more properly restricted to the result of integrating this equation, 
s being obtained explicitly in terms of t. 

119. Units of Mass and Force. — We must first discuss the 
determination of the constant A for certain sets of units. We 
have already spoken of the units of mass and acceleration, but 
since the introduction of motion have said nothing about a 
unit of force. In statics, when we spoke of a force of one pound, 
we meant a force equal to the weight of one pound ; that is, a force 
equal to that necessary to support a mass of one pound and keep 
it from falling. This must then be equal, if we take the surface 
of the earth as fixed, to the force which the earth's attraction 
exerts upon a mass of one pound. If this force is allowed to act 
without any other force being applied, we have the case of a 
freely falling body. It is a matter of experiment that for a given 
locality the acceleration in this case is a constant (for velocities 
small enough that we may disregard air resistance) and is de- 
noted by g. It is equal to about 32 ft/sec^ or 980 cm/sec-. In 
the equation \F = ma, we then have a=:g. Also F — m, for 
the force exerted by gravity on a mass of 1 pound is a force of 
1 pound. Accordingly, if the units are feet, seconds, and pounds 
of mass on one side of the equation and pounds of force on the 
other, A will be equal to g, and the equation becomes gF = ma. 

Problems 
407. If s = ae^*-\-he~^\ show that the particle is acted on by a 
repulsive force which is proportional to the distance from the 
point from which s is measured. 



208 Elementary Mechanics 

408. If a particle moves so that ' 

g — g-ict, (q, gjj^ ht + l) cos ht), 

find expressions for the velocity and the acceleration. Hence 
show that the particle is acted on by two forces, one proportional 
to the distance from the origin and the other proportional to the 
velocity. Describe the motion of the particle. 

120. Falling Bodies. — Within the limits of ordinary obser- 
vation we find that a body allowed to fall freely in a vacuum has 
a constant acceleration g. The value of g varies appreciably 
with changes in latitude and elevation above sea-level. At sea- 
level the value in feet and seconds varies from about 32.09 at 
the equator to about 32.26 at the poles. We may regard 32.2 as 
correct to 3 significant figures in the temperate zone. The 
change with variation in altitude is smaller than this and not 
altogether regular. For certain purposes the value at latitude 
45° and sea-level is regarded as standard. This is about 32.17. 

Frohleins 

409. A body is projected from the, earth vertically with a 
velocity of 40 ft/sec. Find (1) how high it will go before com- 
ing to rest; (2) what time has elapsed when it is at a height 
of 9 feet. 

410. A body moving vertically passes a point at a height of 
54.5 cm. with a velocity of 436 cm/sec. With what initial 
velocity was it thrown, and how much longer will it rise ? 

411. A body moving downward passes a given point with a 
velocity of 50 meter/sec. How long before this was it moving 
upward at the same rate ? 

412. A tower is 288 feet high. One body is dropped from the 
top of the tower and at the same instant another is projected 
vertically upward from its base, and they meet half-way up. 
Find the initial velocity of the projected body and its velocity 
when it meets the descending body. 



Eectilixear MoTioi^r 209 

121. Absolute Units. — The equation XF = ma is of such fun- 
damental importance that it is often desired to simplify it still 
further by a choice of units such that A shall be equal to unity 
and the equation may read F—ma. Of the four units of length, 
time, mass, and force, we might choose any three arbitrarily, 
and then determine the fourth so as to make A=l. This is 
actually done in two principal ways. In both, the units of length 
and time are chosen without reference to this question. Then 
the unit of force is adjusted to that of mass or vice versa. In 
pure science the units of length, mass, and time are usually the 
centimeter, gram, and second (the C. G. S. system). We shall 
then have A = l if we take as a unit of force that force which 
acting on a mass of 1 gram produces an acceleration of 1 cm/sec-. 
This unit of force is called the dyne. It is the principal unit of 
force in all parts of theoretical physics, and is the basis for all 
the electric and magnetic units. The corresponding unit of 
force if we use feet, pounds, and seconds (the f. p. s. system) is 
the poundal, and is that force which acting on a mass of one 
pound produces an acceleration of 1 ft/sec^. These units, the 
dyne and the poundal, are called absolute units of force, as dis- 
tinguished from the pound force and the gram or kilogram force, 
which are called gravitational units because they are equal to 
the weight of the masses having the same names. The term 
absolute is used because the absolute units are the same in all 
localities, whereas the pound force and the gram force vary from 
place to place with the change in the earth's apparent attraction. 

122. Engineer's Units. — Another method of making A=l is 
that sometimes used by engineers. This is to use the gravita- 
tional unit of force and to change the unit of mass. The 
engineer s unit of mass is that mass which acted upon by a 
pound force has an acceleration of 1 ft/sec^. The magnitude of 
this mass may be obtained by setting F and a equal to unity in 

14 



210 Elementary Mechanics 

the equation gF = ma. This gives m=zg; that is, the engineer's 
unit of mass is g times the ordinary unit or pound mass, or ahout 
32 pounds. This unit of mass has no well-acecptcd name, heing 
sometimes called merely the " engineer's unit," sometimes a 
" gee-pound," sometimes a " slug." Most people who use it 
rarely mention its name. 

For many purposes in the use of the engineer's units it is 
not necessary to say whether the value of the pound force is the 
local value or some standard value. For accurate work, however, 
e. g., in the computation of tahles of steam pressure, it is neces- 
sary to define the units more exactly. This may be done by tak- 
ing as a standard pound force the weight of a pound mass at 
sea-level and latitude 45°. The engineer's unit of mass will 
then be the pound mass multiplied by the standard value of g. 
In this case if the actual ciTcct of gravity enters ilie ])roblem it 
must be remembered that it is determined by tlie local value of g, 
and that in general the weight of a pound mass is in this system 
not exactly equal to a pound force. There seems to be no agree- 
ment among engineers on the exact definition of the pound force. 
Some use the local value, others a standard value. There is also 
some disagreement on the definilion of this standard. Fortu- 
nately these discrepancies are in the third or fourth significant 
figure, and hence in many (though not all) practical problems 
they are negligible. 

123. To sum up, we have three principal systems of units of 
force and mass : 

1. The ordinary units, pounds mass and pounds force, or grams 
mass and grams force. 

2. The absolute units, pounds mass and poundals, or grams 
mass and dynes. 

3. The engineer's units, pounds force and slugs. 

In (1) the equation of motion is gF = ma, in (2) and (3) it 

is F—fJla. 



Eectilineak MoTioisr 211 

In (2) both units are constant. In (1) the unit of mass is con- 
stant; the unit of force is the local pound and depends on the 
local value of g. 

In (3) the practice of engineers varies, some using a local 
pound force and a mass unit to agree with it, others a standard 
pound force and a corresponding mass unit. 

The units in ( 1 ) are those regularly used in every-day speech, 
those in (2) are regularly used by physicists. It must be under- 
stood that by no means all engineers use (3). Some use (1) 
and electrical engineers almost universally use (2). 

The student should be prepared to understand statements and 
problems in any one of these systems of units^ as he is likely to 
meet all of them in practice. He should also be able to convert 
any one into the other. 

Example 1. — Show that a poundal is about equal to the weight 
of half an ounce. 

Solution. — Since the result is wanted in ordinary units, we 
use the equation gF = ma. From the definition of a poundal a=l 
when m = l. Hence gF—l, and F—l/g in ordinary units, i. e., 
3^2 of a pound or -| ounce. 

Example 2. — How many dynes are in a force equal to the 
weight of 1 gram ? 

Solution. — As in the preceding, a force of 1 dyne is 1/g in 
ordinary units, i e., 1/980 of a gram; or a gram weight equals 
980 dynes. 

Problems 

413. Find the magnitude in dynes of the force which, acting 
on a kilogram for 5 seconds, produces in it a velocity of 1 
meter/sec. 

414. A force equal to the weight of a kilogram acts on a body 
continuously for 10 seconds and causes it to describe 10 meters 
in that time. Find the mass of the body. 

415. A body of mass 200 tons is acted upon by a force of 
112,000 poundals. How long will it take to acquire a velocity 
of 30 mi/hr ? 



212 Elementary Mechanics 

124. Constant Force. — Example 1. — A mass of 10 pounds is 
acted on by a constant force of 25 pounds. If it starts with a 
velocity of 20 ft/sec in the same direction as the force acts, how 
far will it travel in the first three seconds ? When will it have a 
velocity of 400 ft/sec ? 

Solution. — As the problem is stated in ordinary units, we have 
gF = ma where g — 32, F = 25, m = 10. Hence a = SO = dv/dt. 
This is all there is of dynamics in the problem; the rest consists 
in integrating and determining constants. Thus v = SOt-]-C. 
Since 7; = 20 when ^ = 0, we have C = 20, and v :=80t-\- 20 = ds/dt. 
Hence s = 40^- + 20^ + Ci; but C^ — because we measure the 
distance naturally from the starting point, i. e., take 5 = when 
^ = 0. The equation of motion of this body is' s = 40^- + 20f. 
When t = 3, s = 420, i. e., the body travels 420 feet in the first 
three seconds. To find when it has a velocity of 400 ft/sec put 
i; = 400 in the equation connecting v and t; then 400 = 80^ + 20, 
if = 4.75. Hence the body will take 4.75 seconds to acquire a 
velocity of 400 ft/sec. 

Example 2. — Show that the distance traveled by a body start- 
ing with a velocity Vo and having a constant acceleration a, is 
equal to the algebraic sum of two distances^ one the distance it 
would travel with a constant velocity Vo, and the other the dis- 
tance it would cover if it started from rest with the acceleration a. 
Determine, by integration, the constants for each of the following 
three cases: 

(1) Motion with constant velocity Vo. (2)* Starting from 
rest with acceleration a. (3) Starting with velocity Vo and hav- 
ing acceleration a. 

Example 3. — A mass of 40 grams is acted on by a constant 
force of 100 dynes. If it starts with a velocity of 250 cm/sec in 
the direction opposite to that in which the force acts, how far 
will it have gone when its velocity is reduced to 150 cm/sec? 
How far when it comes to rest ? 

Solution. — Using absolute units, the equation F = ma becomes 
-100 = 40a, a=-2i, i;- = i)o' + 2a5, or (150)2={250)2- 
2(2^)5, or s = 8000 cm. In the second case v = and s = 12500 
cm. Note that the equation v^ = Vo'^ + 2as does not contain t, 
and is therefore particularly useful if t is neither given nor 
required. 



Eectilinear Motion" 218 

Prohle7ns 

416. A mass of 1-0 pounds falls 10 feet from rest, and is then 
brought to rest by penetrating 1 foot into some sand. Find the 
average force of the sand on it. 

417. A bullet with initial velocity 1500 ft/sec strilves a target 
1200 yards away with velocity 900 ft/sec. The path being sup- 
posed horizontal, find the ratio of the mean resistance of the air 
to the weight of the bullet. 

125. In many problems it is possible to avoid altogether the 
question of units of mass. If in any system a force F acting on 
a body of mass m produces an acceleration a we have 

XF=ma. 

If a new force F' acting on the same body produces an acceler- 
ation a, we have in the same system of units 

XF' = ma. 

Dividing one of these by the other 

F[_a 
F' a'' 

If now we take the special case where the force F' is the weight 
of the body W, a will be g, the acceleration of gravity, and we 
have 

W-g' 
Here it is only necessary that the weight W shall agree with F 
as regards the units in which it is measured and agree with g as 
regards the locality in which it is taken. This equation, written 
in the form 

gP=Wa, 

is substantially the same as that gF — ma used in the system (1), 
if we remember that in that system mass and weight are numeri- 
cally equal. 



214 Elementary Mechanics 

Pro'ble7ns 

418. A force, acting on a mass of 10 pounds for 5 seconds, 
produces in it a velocity of 100 ft/sec. Compare the force with 
the weight of a pound, and find the acceleration it would produce 
in a ton. 

419. A bullet moving 200 ft/sec is fired into a block of wood 
into which it penetrates 9 inches. If a bullet moving with the 
same velocity were fired into a similar piece of wood 5 inches 
thick, with what velocity would it emerge, supposing the resis- 
tance to be uniform ? 

420. A ball of mass 1000 grams is discharged with a velocity 
of 45000 cm/sec from a gun the length of whose barrel is 200 cm. 
Find the mean force exerted on the ball. 

126. Relative Acceleration. — It should be observed that the 
value of g that we have spoken of so far is an acceleration meas- 
ured relative to the surface of the earth taken as fixed. The 
acceleration of a falling body if referred to some other body 
regarded as fixed will be different, if the latter has an acceleration 
with respect to the surface of the earth. Thus if the supporting 
cable of an elevator should break, allowing the car to fall freely, 
the car and everything in it would have an acceleration g with 
respect to the earth. The acceleration of an unsupported body 
within the car would be g if referred to the earth, but zero if 
referred to the car. Thus an unsupported body w would remain 
at the same distance from the floor of the car as long as the car 
had the downward acceleration g, unless it were given an initial 
velocity with respect to the car. It is clear then that a body 
within the car would entirely lose its weight if suspended and 
weighed by an apparatus attached to the car. 

Example 1. — A man weighing 160 pounds stands on the plat- 
form of a spring scale in an elevator. If the elevator starts up 
with an acceleration of 4 ft/sec^, what will the scale register ? 

Solution. — The man^s body is acted on by two forces, his 
weight of 160 pounds, and the upward pressure P exerted by the 
scale platform. The equation of motion then becomes 
^(P-160) =160x4, or if ^ = 32, P = 180. 



Eectilinear Motion 215 

Example 2. — Show that if the weighing apparatus is subject 
to any upward acceleration, the apparent weight of a body is 
increased by an amount which bears the same ratio to its weight 
at rest as the upward acceleration bears to the acceleration of 
gravity. Does this statement apply to a beam balance? 

Solution. — Let the student proceed as in the foregoing example, 
using general values in terms of letters for each of the quantities 
involved. The statement does not apply to a beam balance, be- 
cause the weights in both scale pans would be increased equally 
by the acceleration. 

Problems 

421. A man whose weight is 112 pounds stands in an elevator 
which moves with a uniform acceleration of 12 ft/sec". Find the 
force he exerts on the floor of the elevator when it is (1) ascend- 
ing, (2) descending. 

422. A balloon ascends with uniform acceleration so that a 
mass of 100 pounds exerts a force of 116 pounds on the floor of 
the balloon. Find the height of the balloon one minute after 
starting. 

127. Motion on an Inclined Plane. — If a body rests on an in- 
clined plane, and is not acted on by any force having an upward 
component sufficient to lift it off the plane, any motion that it 
may have will keep it in contact with the plane. If the forces 
and the initial velocity are all in a vertical plane perpendicular 
to this one, the motion will be rectilinear. In all cases of this 
sort the body is acted on by the force of gravity and also by the 
reaction or upward pressure of the plane, together with any other 
forces that may be present. If the plane is perfectly smooth, the 
reaction of the plane is normal to its surface. If it is not smooth 
the reaction may be resolved into two components, one normal 
and the other along the surface. Since the body remains in con- 
tact with the plane, no motion takes place in the direction normal 
to the surface, and hence the normal components of all the forces 
acting must be in equilibrium. If the other forces are known 
this determines the normal reaction of the plane, and from this 
we can determine the force of friction, if any. 




316 Elementary Mechanics 

Example 1. — A body is projected with an initial velocity of 
80 feet per second up a perfectly smooth plane inclined at an 
angle of 30° to the horizontal. (1) How far from the starting 
point will it be at the end of 4 seconds? (2) How far up the 
plane will it go? (3) When will it return to its starting point? 
(4) What velocity will it then have? (Take ^ = 32.) 

Solution. — If the force of gravity W is resolved into two com- 
ponents normal to the plane and along 
it (Fig. 197) the former is balanced 
by the normal reaction of the plane, N : 
the latter, W sin 30% or iW, is the 
resultant force acting down the plane. 
Fig. 197. The equation gF=Wa then becomes 

— ^Wg = Wa or a = — -i-pf = — 16. Hence 
v = 80-16t, and 5 = 80^-8^"^ At the end of 4" seconds 5 = 80 x 
4 — 8x16 = 192 feet. The second question is to determine the 
maximum value of s. This of course occurs when its derivative v 
is zero. Then = 'i;o2 + 2as, 5= -V/2a= 6400/32 = 200 feet. The 
third question is to determine t when 5=0 for the second time. 
This gives S0t — 8t^ = 0, t — O, and t — lO, the second root being the 
one required. The fourth question is answered by putting this 
value of t in the expression for v. This gives v = 80 — 160 = — 80. 
The velocity is thus equal to the initial velocity, but in the opposite 
direction. 

Example 2. — Show that if a body is projected with any initial 
velocity up a perfectly smooth plane, it will pass any point on 
the downward journey with the same speed that it passes the 
same point going up. 

Solution. — Form the equation of motion with the general ini- 
tial velocity v^ and the inclination of the plane 6 and then use 
the equation connecting v, a, and s in uniform acceleration. 

Example 3. — Solve Example 1 with the plane having a coef- 
ficient of friction /;t = 0.50. 

Solution. — The student should first note that the coefficient of 
friction is supposedly known to only two significant figures. (A 
coefficient of friction is very rarely known any more accurately 
than this.) Consequently there is no object in taking a value 
of g more accurate than this, i. e., g should be taken as 32 rather 
than 32.2 or any more accurate value without specific directions. 



Eectilinear Motion 



217 



Let the student also note that unlike Example 1 the equation of 
motion is not the same for the downward motion as for the up- 
ward. There the only force acting along the plane is a component 
of the weight. Here there is also a force of friction and this acts 
downward when the motion is upivard and upioard tvhen the 
motion is doivnward. Hence the resultant force and the conse- 





FiG. 198. 



quent acceleration are different in the two cases (Fig. 198). For 
the upward motion 

F=-W sin 30^- /.T7 cos 30° = ]F( -i-iV^), 
a = (7(--i-|V3) = -.93^=-30, 
^ = 80-30f, 
5 = 80^-15^^ 

Now we must observe that we can not answer question (1) by 
putting ^ = 4 in this last equation, unless we make sure that this 
equation of motion continues to hold for 4 seconds. That it 
actually does not hold when t = 4, we can readily see by finding 
V at that instant and observing that it has become negative. 
This equation holds then only until v — Q or t = 2.7. We can, 
however, answer question (2) at once by the equation s = Vo^ /2a = 
6400/60 = 107. The new acceleration is ai = ^(-i + iV'3) = -2. 
The time reckoned from the instant of rest is 1^ = ^ — 2.7 = 1.3. 
Then s = 107 + iai^i2 = 107-/i- = 107-1.7 = 105 as accurately as 
our data warrant. To determine the time to return to the start- 
ing point we have s = = 10 7- t^" or ^i = 10. 3, / = 10.3 + 2.7 = 13 
seconds. Also v^^ = V2^= V - 4 x - 107 = 21 f /s. Thus the 
body takes 10.3 seconds to come down the distance that it went up 
in 2.7 seconds, and acquires a velocity of only 21 f/s instead of its 
original velocity of 80 f/s. 



218 Elementary Mechanics 



Prohlemi 



423. A body is projected with a velocity of 80 ft/sec up a 
smooth inclined plane whose inclination is 30°. Find the space 
described and the time that elapses before it comes to rest. 

424. A particle slides without friction down an inclined phme, 
and in the fifth second after starting passes over a distance of 
2207 cm. Find the inclination of the plane. 

425. A particle slides down a rough plane inclined to the hori- 
zontal at an angle a. If fx be the coefficient of friction, find the 
acceleration and velocity when the body has moved a distance /. 
. 426. One body slides down a smooth inclined plane and another 
drops vertically to the same level. Show that the times are pro- 
portional to the spaces described, and that the velocities acquired 
are equal. 

128. Atwood's Machine. — We have been considering so far 
the motion of a single body. We shall now consider the related 
motions of two or more bodies which are connected in such a 
way that when we know the acceleration of one we know the 
acceleration of the others. The simplest case is that of two 
bodies directly connected by a weightless, flexible, inextensible 
cord, which remains taut during the motion, and the part ad- 
jacent to each body lies in the rectilinear path of that body. Two 
weights connected by a cord passing over a pulley, the weights 
being otherwise unsupported, constitute aii Atwood's machine. 
The pulley is made as light as possible, and other- 
wise designed to offer as little resistance as possible 
to motion and acceleration. In a first approxima- 
tion any resistance of the pulley may be negleetecl. 
We may then solve problems connected with an 
Atwood's machine if we observe the following facts : 
(1) Newton's Second Law applies to each hodij 
separaiely if we use the total resultant force acruuj 
on that body; (2) The forces acting on each body are its weight 
and the tension in the cord; (3) Newton's Third Law tells us that 




Eectilixear Motion- 219 

the tension in the cord (neglecting the pulle}- resistance) is the 
same as applied to each body, but draws one forward and the other 
backward; (4) the acceleration is the same for the two bodies if 
that of each is taken in its own path. 

Example 1. — Tlie weights on the two sides of an Atwood's 
machine are 22 pounds and 2G pounds. Find (1) the distance 
the weights will travel in 3 seconds starting from rest, (2) the 
tension in the cord, (3) the force necessary to support the pulley 
during the motion. 

Solution. — Applying the principles just enumerated, the equa- 
tions of motion for the two bodies are (7(26 — T) = 26a, and 
g{T — 22) =22a, where T is the tension in the cord. Note that 
the positive direction is the direction of motion in each case, as 
only in this way do we have the same a in the two equations. In 
equations of this type T may always be eliminated by direct 
addition. This gives us -i^ = 48a = 128, a = S/3 f/s". Then 
s=:^at- = 12. Hence the weights travel 12 feet in the first 3 
seconds. Putting the value of a in the original equations enables 
us to find T and check it. This gives T = 23^ pounds. The 
pulley is acted on by two downward forces each equal to T. The 
supporting force therefore is equal to 2T or 47f pounds. 

Example 2. — If the weights on an Atwood's machine are 
10.000 kilograms and 10.574 kilograms, they move 190.2 cm. in 
exactly 4 seconds, starting from rest. If one weight is 2.000 
kilograms it is found that the other must be 2.176 kilograms to 
produce the same motion as before. Find the value of g. 

Solution. — The two observations enable us to eliminate the 
resistance of the pulley if we merely assume that the frictional 
resistance is sensibly constant with the change of load. The 
angular acceleration of the pulley is actually the same in the two 
cases. If we take account of the resistance of the pulley, the 
tensions in the two parts of the cord are different, the difference 
being the force necessary to produce the motion of the pulley. 
If we call these in the first trial T^ and T^ we have 

5r(10.574-ri)=10.574a, 
^(T/-10.000)=10.000a. 



220 . Elementary Mechanics 

Adding these, we have 

^(0.574-T, + r/)=20.574a. (1) 

If To and T^ are the tensions in the second trial, we shall have 
similarly 

^(0.176-To + r/)=4.176a. (2) 

But T^ — T^ = To — ^2',' as each is the force required to move the 
pulley. Hence if we subtract (2) from (1) we have 0.398^ = 
16.398a. Buts = ia^- = 8a = 190.2, a = 23. 78, ^ = 16.398x23. 78-- 
0.398 = 980, or if the indicated weights are correct to tenths of a 
gram ^ = 979.5 cm/sec.^. 

Prohlems 

427. A mass of 9 pounds descending vertically drags up a mass 
of 6 pounds by means of a string passing over a smooth pulley. 
Find the acceleration of the mass and the tension of the string. 

428. Two equal masses, m, are connected by a string passing 
over a smooth pulley. What mass must be taken from one and 
added to the other so that each mass may describe 200 feet in 
5 seconds? 

429. Two masses of 3 pounds each are connected by a light 
string hanging over a smooth peg. If a third mass of 3 pounds 
be laid on one of them, by how much is the pressure on the peg 
increased ? 

430. A mass of 4 ounces is attached by a string passing over a 
smooth pulley to a larger mass. Find the magnitude of the latter 
so that, if the string he cut after the motion has continued for 3 
seconds, the former will ascend 16/9 feet before descending. 

129. Other Problems on Connected Bodies. — The Atwood's ma- 
chine is only one case out of many of connected bodies. If two 
bodies are connected by a cord which lies in their line of motion, 
the two must have the same acceleration. If the connection is 
made by means of pulleys, we can easily see the relation between 
the distances the bodies travel, and hence between their acceler- 
ations. If one of the bodies rests on a horizontal or an inclined 
plane, the pressure of the plane must be taken account of, and 



Rectilinear Motion 



221 



friction if any. Except for pulley resistance, which we shall 
usually neglect, the tension in a cord is the same throughout. 

Example 1. — A weight of 20 pounds rests on a smooth table 
(Fig. 200). A cord fastened to 
it passes over a smooth pulley and 
is attached to a weight of 40 
pounds otherwise unsupported. 
Another cord fastened to the first 
weight passes over a smooth pulley 
at the opposite edge of the table 
and is attached to a weight of 10 







20 


T, 






•^ y^ 




>- 




<P^ 


/ 







^ 


^ 








40 



Fig. 200. 



pounds which rests on a smooth 

plane whose slope is f . Find the acceleration of the system if 

^ = 32.2. 

Solution. — If T^ is the tension in the first cord and T^ that 
in the second, the equations of motion of the 10, 20, and 10 
pound weights are respectively 

(7(40-ri)=40a. 

,g(Ti-To)=20a, 

^[2^-f(10)] = 10a. 

Adding these three equations, we have 

^[40-f(10)] = T0a, 
or 

a= -^^^^^ =rl5.7f/s2. 

Example 2. — If the coefficient of friction between the 10- and 
20-pound weights and their planes in the preceding example is 
0.30, find the acceleration of the system. 

Solution. — The three equations of motion will now be 

^(40-Ti)=40a, 
^[ri-T.-20(0.30)]=:20a, 
and 

^[^2-1(10) -f (10) X(0.30)]=10a. 
Adding these, 

^(40-6-6-2.4) =25.6(7 = 70a. 
Hence 

a = 11.8 ft/sec^. 



222 



Elementary Mechanics 



Example 3. — One end of a cord (Fig. 201) is fastened at the 
top of an inclined plane (inclination 30°). The cord passes 
over a simple pulley attached to a weight of 100 pounds resting 
on the plane (/x = i) and then passes back over a fixed pulley at 
the top of the plane and is attached to a weight of 80 pounds 
otherwise unsupported. Neglecting the pulley resistance, find 
the acceleration of the 100-pound weight and the tension in the 
cord. 




Fig. 201. 



Solution. — Let a be the acceleration of the 100-pound weight. 
Then that of the 80-pound weight will be 2a. The total force 
acting to advance the 80-pound weight is 80 — T, where T is the 
tension in the cord. The forces acting on the 100-pound weight 
are a tension T in eacli part of the cord, the component of the 
weight down the plane, 100 sin 30°, and the friction, -|(100) 
cos 30°. The equations of motion of the two bodies will then be 



and 



^(80-T)=80(2a) 
^(2r-100 sin 30° -50 cos 30°) = 100a. 



Eliminating T, we have 



^(lGO-100 sin 30° -50 cos 30°) = 420a, 
„ 110-25A/3^^5_^f^/^^^,^ 



420 



From the first equation, 
160a _ 
9 



T = 80- 



80 



160(110-25V3) _ 



420 



= 54 lbs. 



Eectilinear Motion 223 

Problems 

431. A mass of 12 pounds drags a mass of 16 pounds up a 
smooth plane of inclination 30°, the masses being attached by a 
string passing over the top of the plane. Find the distance 
described in 5 seconds and the tension of the string. 

432. A mass of 6 ounces slides down a smooth inclined plane 
whose height is half its length, and draws another mass from rest 
over a distance of 3 feet in 5 seconds along a smooth horizontal 
table which is level with the top of the plane over which passes the 
string which connects the two masses. Find the mass on the 
table. 

433. A mass of 5 pounds on a rough horizontal table is con- 
nected by a string to a mass of 8 pounds which hangs over the 
edge of the table. Find the coefficient of friction in order that 
the heavier mass may move vertically with half the acceleration 
it would have if it fell freely. 

130. Motion Under a Variable Force. — So far avc have dis- 
cussed only motions in which the force and the resulting acceler- 
ation are constant. In many practical problems, however, the 
force varies. It may depend on the time, on the distance, on the 
velocity, or on any combination of these. The case of the force 
depending only on the time is the easiest to handle; but it is of 
very rare occurrence in practice. It will be included for the sake 
of completeness. In all problems on vibrations, the force de- 
pends on the distance, i. e., on the displacement of the body 
from its mean position. It depends on the velocity in all prob- 
lems involving fluid friction or fluid resistance. The possi- 
bilities of combination should be obvious. For example, in 
interior ballistics the force propelling the projectile evidently 
depends on the time and the distance, — very slightly if at all on 
the velocity. In exterior ballistics, however, the force retarding 
the projectile depends only on its velocity, not at all on the time, 
nor in ordinary cases on the position. The wide range of appli- 
cation is suggested by the few illustrations mentioned. It should 
also be kept in mind that a vast number of problems in all parts 



224 Elementary Mechanics 



of Ph3^sics are solved in a manner identical with these in 

Mechanics. In all of these problems we may write a~~^F, 

where F is a function of t, s, or v, or a combination of them. We 

may then substitute for a either --r^, -=— , or ^ ? ^^ ^^^y ^^st 

suit the problem. In nearly all cases g/W will be merely a con- 
stant factor. (A rare exception to this last occurs in the case of 
a moving body such as a rocket which loses part of its substance 
as it moves. In this case W is variable.) 

131. Force Depending on the Time. — Writing F—F{t), we 

have 

-^-^'^^^ 
v^-^SF{t)dt + C, 

s=\vdt + C^. 

In order to determine C and Cj, we must know two " initial con- 
ditions " or the like, i. e., the values of v and s for a particular 
value of i, or the values of s for two different values of t, or two 
other similar conditions. 

Example 1. — A mass of 8 pounds starts with a velocity of 
2 ft/sec and is acted on by a force of 6t^ — 4:t pounds in the 
direction of its initial velocity, where t is in seconds. Find the 
distance covered in 3 seconds and the velocity at the end of that 
interval if ^ = 32. Here 

But v = 2 when ^ = 0. Hence C = 2 and 

t;==% ==8/^-8^2 + 2. 
dt 

Integrating again, 

s = 2t^- it^-^2t. 



Eectilinear Motion . 225 



the constant of integration here being zero, becanse 5 = when 
t = 0. By putting ^==3 in the fornnihis for s and v, we have that 
at the end of 3 seconds the body is 96 feet from its starting point 
and has a forward velocity of 146 ft/sec. 

Example 2. — A moving body is observed to pass a fixed point 
at a certain instant. It is observed again at the ends of. 2, 4, and 
6 seconds to be 202, 368, and 486 cm. respectively from this 
point. If it is subjected to a retarding force which is propor- 
tional to the time elapsed since the instant the force is first 
applied, determine (a) the original velocity, (b) when and 
where the force was first applied, (c) when and where the body 
comes to rest, (d) the mass of the body in grams if the retarding 
force when it comes to rest is 250 dynes. 

Solution. — Let to be the instant when the force is first applied. 
Then the force and consequently the acceleration will be pro- 
portional to t — to, or 

a = l{t-t,), (1) 

Integrating this twice, we have 

v=.ll,{t-t,y + v,, (2) 

and 

s = \lc{t-t,Y + v,t-^C, (3) 

where Vo is the original velocity when the force is apphed. If .9 
is measured from the first observed point, we have for deter- 
mining the four unknown constants, h, t^, v^, and C, the four 
values of s, namely, 0, 202, 368, and 486, corresponding to ^ = 0, 
2, 4, 6, respectively. Putting these in (3), 

= iA;(-^o)^ + C; (4) 

202=:iA;(2-^o)^-f2i'o + C; (5) 

368 = l^^(4-^o)3-|-4^'o + C; (6) 

486 = i^'(6-^o)' + 6i'a + C. (7) 

Subtracting (4) from each of the succeeding, and multiplying 

by 6, 

1212 = A-(8-12^o + 6fo')+12i'o; (8) 

2208 = ^^(64-484 + 12^o')+24i;,; (9) 
2916 = ^(216-108)50 + 1842) +36?;o. (10) 



226 • Elementary Mechanics 

Eliminating v^ between (8) and each of the others, 

-216 = ^'(48-24^o); (11) 

-720 = A:(192-72^o). (12) 

Finally, eliminating to, 

-72 = 48^% A;=-f. 

Substituting successively in (11), (8), and (4), we get 

to=-4:, t;o = 120, and = 16. 
Hence the equation of motion is, from (3), 
s=-i(^ + 4)3 + 120^ + 16. 

AVhen ^ = /„=— 4, 5=— 464. Hence, answering (b), the force 
was applied 4 seconds before the first observation, and at a point 
464 cm. from the first observed point. Substituting in (2), 

t;rr-|(^ + 4)2 + 120. 

When the body comes to rest v = 0, and hence 

^=VT60-4 = 8.65 and 5= -J; (160) -^ + 120(8.65) +16 = 548. 
These results answer (c). Substituting in (1), 

a=-f(^ + 4). 

When t = 8.65, this gives ar= -18.97 and 
F -250 



18.9^ 



= 13.2 o-rams. 



132. Force Depending on the Distance. — In this case it is 
always advisable to express the acceleration in the form 

dv dv 

dt ds 

We then have a first-order differential equation connecting v 
and s. If w^e can integrate this we have an equation (without 

derivatives) between v and 5. Then putting v— ^- we have to 

integrate another first-order equation, in s and t, in order to 
finish the solution of the problem. 



Eectilixear Motiojq" 227 

"We should bear in mind the importance of the correct deter- 
mination of the constants of integration. This is done by the 
same methods as those used in the examples of the preceding- 
article. 

Example 1. — A body when displaced from its position of equi- 
librium is acted on by a force tending to restore it, which is pro- 
portional to the amount of the displacement. If this force is 
40 pounds when the displacement is 2 feet, determine the motion 
of the body if it weighs 160 pounds and starts from rest with a 
displacement of 4 feet. 

Solution. — If s denotes the distance of the body from its 
position of equilibrium, then F is proportional to sj that is, 
F = lis. To determine l- we have that F=—4,0 when s = 2, the 
sign of F being negative because the force tends to decrease s. 
Hence A-= —20. and F= — 20s. Xote that this makes F negative 
when s is positive and positive when s is negative, and thus 
agrees with the physical fact that the force always tends to re- 
store the body to the position of equilibrium, and hence acts in 
the direction opposite to the displacement. The equation of 
motion is then 

gF=-20gs = 160a; 

^^ ds ' 

2v dv= —^gsds; 
V^=-Xgs--\-C. 

To determine C we know that v = when s = 4. Hence C = 2g and 

Before performing the second integration we may infer sev- 
eral characteristics of the motion from this equation in v and 5. 
Thus, as regards numerical values, v decreases as s increases; 
becomes zero when s reaches a maximum, and reaches a maximum 
when 8 = 0. Also at any distance s from the mean position the 
value of V- is determined, but v may be positive or negative. This 
shows that the body passes any given point with the same speed 
whether going or returning. 



228 Elementary Mechanics 



If we continue the integration^ we have 
ds 



•=i = ^|i■^ 



'=^J]lvfo = ^l^■»-i-''■• 



When t = 5 = 4; hence 
and 

Solving for s, we have 

s = 4eos^|t 

This shows us, in addition to the information we had before, 
that the motion is periodic; that is, it exactly repeats itself after 

a certain time interval. For the trigonometric function cos ^ |^ 
will return to the same value if -^l^t is increased by 27r; that is, 

if t is increased by 27r ^ - . This quantity is called the period 

of the motion. The maximum displacement, in this case 4 feet, 
is called the amplitude. 

133. Simple Harmonic Motion.~The preceding example is 
an illustration of Simple Harmonic Motion. This is defined in 
general as motion in which the force is proportioiial to the dis- 
placement of the body from its niean position and acts in a 
direction to restore it to that position. From the equation of 
motion it follows that the acceleration is equal to a negative 
constant times the displacement. If this constant is called —k^, 
we have a— —h'^s. A complete solution of this differential 
equation is 

5=^ ^mh{t-to), 



Eectilixear Motion 229 

where to and Vo are the time and the velocity when 5 = 0. The 
period of this motion is ^tt/I: and its amplitude is Vo/1^'. The 
quantity to, which determines when the motion starts, is called 
the phase. Frequently instead of the phase, the phase angle is 
spoken of. This is Hq and is often given in degrees instead of 
radians, especially in the treatment of alternating currents, 
which is very closely related to this subject in Mechanics. 

ProhJems 

4:34:. Integrate the equation a=—l'-s and obtain the general 
equation of simple harmonic motion with the constants expressed 
in terms of to and a, the phase and amplitude. (The method of 
solution is precisely the same as in the numerical example which 
was worked out.) 

435. Show that the period of a simple harmonic motion depends 
only on the force of restitution and not at all on the initial 
conditions. 

436. Express the maximum speed in terms of the amplitude. 

437. Show that in any simple harmonic motion the acceler- 
ation and displacement reach numerical maxima when the velocity 
is zero, and are equal to zero when the velocity is a luaximum. 

438. Find the period, amplitude, phase angle, and maximum 
velocity for the motion 

s = 3 sin (4^ — tt). 

439. By expanding sin A'(f — /q) as the sine of a sum show- 
that any simple harmonic motion can be written in the form 

5 = .4 sin l-t-\-B cos li, 

and express .4 and B in terms of amplitude, period and phase. 

440. A certain spring is stretched one inch by a force of 10 
pounds. A weight of 20 pounds is hung on the spring in its 
unstretched condition. Assuming the spring obeys Hooke's Law, 
determine the amplitude and period of the resulting motion. 

441. Write the equation of a simple harmonic motion having 
the amplitude 5 feet, period 3 seconds, and phase angle 60°. 

442. Below the surface of the earth the force of gravity is 
directly proportional to the distance from the center. If a hole 
could be bored straight through the earth, and a body were 



230 Elementary Mechanics 

dropped in, how long would it take to reach the antipodes and 
what wonld be its maximum velocity? 

134. Force Depending- on the Yelocity.— Example. — A body 
whose motion is retarded by a lliiid friction has an acceleration 
— lev where v is the velocity. Find the distance covered in time i 
if the initial velocity is Vo. 

Solution. — 

y dv 

V 

V 

— M = \ogv + C = log V — log Vo = log ■ ; 



IC K 

if 5 = when / = 0. 

Prohlcms 

443. Show tliat ilie distance covered in tlie preceding example 
can never exceed Vo/h'. 

44-4. Solve the same problem if the retarding force is pro])or- 
tional to the square of the velocity. 

445. In this last problem is there (theoretically) any upper 
limit to the distance covered? 

Revietv Prohlems 

44G. A 10-pound body moving 4 ft/sec meets a 12-pound body 
moving 7 ft/sec in the opposite direction. If they unite into one 
body, find its velocity. 

447. An Atwood^s machine, with suspended masses m^ and 
mo, is placed on the platform of a set of scales. Find the change 
in its apparent weight when the masses are allowed to move. 

448. A string hung over a pulley has at one end a mass of 10 
pounds and at the other end two masses of 8 and 4 pounds respec- 
tively. After being in motion for 5 seconds the 4-pound mass is 
taken off. Plow much further will the masses go before they 
come to rest? 



Eectilinear Motion 231 

449. A train of mass 200 tons is runnino^ at the rate of 40 
mi/hr down an incline of 1 in 120. Find the resistance neces- 
sary to stop it in half a mile. 

450. A car starting from rest coasts for 1 mile down an incline 
of 1 in 100. If the resistance be equal to 8 pounds per ton, how 
far will the car be carried along the horizontal level at the foot 
of the incline? 

451. A stone is dropped into a well and the sound of the splash 
is heard in Ty^Q- seconds. If the velocity of sound is 1120 ft/sec 
how deep is the well? 

452. A cage in a mine shaft descends with 2 ft/sec- acceler- 
ation. After it has been in motion for 10 seconds, a particle is 
dropped on it from the top of the shaft. When will the particle 
hit the cage ? 

453. A particle of unit mass moves in a straight line so that 

s = 6 — 5 sin- ^, where t is the time and 5 the distance from a 

point 0. Find when the particle is moving forward and when 
backward. Find also the greatest distance which the particle 
reaches from and the force which acts upon it. 

454. A particle falls from height li upon a horizontal plane. 
If G is the coefficient of restitution, find the whole distance de- 
scribed by the particle, and the time that elapses before it has 
finished rebounding. 

455. The masses of 5 balls at rest in a straight line form a 
geometric progression whose ratio is 2 and their coefficients of 
restitution are each f . If the first ball be started toward the 
second with velocity v, find the velocity communicated to the 
fifth. 

456. Two scale pans, each of mass 3 pounds, are connected 
by a string passing over a smooth pulley. Show how to divide a 
mass of 12 pounds between the two pans so that the heavier may 
descend a distance of 50 feet in the first five seconds. 

457. Two strings pass over a smooth pulley; on one side they 
are attached to masses of 3 and 4 pounds respectively and on the 
other both are attached to a single mass of 5 pounds. Find the 
tensions of the strings and the acceleration of the system. 

458. A mass m pulls a mass m' up a smooth plane of incli- 
nation ^, by means of a string passing over a pulley at the top 
of the plane. Find the acceleration of the system. 



232 Elementary Mechanics 

CHAPTEE IX 
WORK AND ENERGY 

135. Work. — When a force acts on a fixed point of a body, 
and the point of application moves so that the displacement, 
however small, has a component in the direction of the force, the 
latter is said to do work. Thus a force does work on a particle 
when and only when the particle moves nnder the action of the 
force. Force and distance are essentials of work. Force, how- 
ever long it may act, which produces no motion, does no work. 
For example, a weight of 100 pounds rests on a rough horizontal 
plane for which />t = i. A horizontal force of 15 pounds is ap- 
plied. The body will not move, since the force of friction 
(which may be as great as 25 pounds) will be just sufficient to 
produce equilibrium. According to the mathematical definition 
of work, the 15-pound force will do no work, however long it 
acts. This also illustrates that time is not an essential element 
of work, as mathematically defined. Furthermore, no work 
results from motion without force. For example, the resultant 
work done on a particle moving with uniform velocity, in a 
straight line is zero. An unbalanced force always does work, for 
it always produces motion. 

136. Work Done by a Constant Force. — If a constant force F, 
acting on a particle, displaces it a distance s along the line of 
action of the force, the work done on the particle by the force is 
defined by the equation 

Work = i^Xs. 

That is, the work is the product of the force and the displace- 
ment of the particle. 

It appears then that the unit of work involves the unit of 
force and the unit of length. For example, if a force of A 



Work and Energy 233 

pounds ju'ts Ihrouu'h a div^taiu'o o( /> foot or a foroo o( /> pounds 
acts throui^h a distanoo of .1 fool, tlio work dono is AH foot- 
pounds. Tho wcu'k is posit ivo whon tho disphu'onuMit is in tho 
dirooiion of tho (oww and no^ativo whon tho displaoomont is in 
the dirootion opposito to that oi' tho fori'o. In tho fornior oaso work 
is said to bo dono b}' tlio foroo, whik^ in tho lattor work is said to 
be dono against tho foroo. Tlius, tho work dono by tlio foroo of 
gravity on a body woighing "JO }Hnn\ds, whioh falls 10 foot, is 
:200 foot-pounds. Tho work dono against tho foroo of gravity by 
a foroo whioh lifts a '?0-pound body 10 foiM is "^OO foot-pounds. 
The hittor may also bo oxprossoil by saying tho work dono by tho 
foroo of gravity is — "JOO foot-pounds. 

E.vaniplc 1. — A body of woight o^.'2 pounds uioves in a straight 
line with uniformly aoeoloratod motion. If tho aoooloration is 
10 feet per second per second , lind tho work dono on tho partic-lo 
in moving it a distance of "Jo foot. 

Soluiion. — From Newton's sooond law of motion, 

F = il/tt=: '!*^'; 10 = 10 lbs. 

From tho dolinition of work wo have, 

Work = foroo \ displaoomont — v\^0 ft. -lbs. 

E.vaniplc :3. — A train weighing Tu^tl ions is nuning along a 
straight level track when the power is suddenly shut olV. It is 
then brought to rest in 10 seconds by the friotion o[' the brakes. 
If the friotion is 10 pounds per ton, tind the work dono by the 
friction. 

Soluiion. — From Newton's second law of motion, 

,00X10=— -^--^. a, 



wlionce 

52^ 



a= ^-y-^r ^^' P^'^' ^*-'^'- P^'^" ^*-'^'' 



234 Elementary Mechanics 



The distance the train moves before coming to rest is then 
^ ,. 1 32.2 .-.^.o 1610 ., 
^ = '"^'==2^2r(^')"=-22l ^'- 
From the definition of work we have^ then, 

VVork = force times displacement = 35,900 ft. -lbs. 

Problems 

459. A body weighing 100 pounds rests on a rough horizontal 
plane (/x = 0.5). (a) What work is done in moving the body 10 
feet in any direction along the plane? (b) Has the time it takes 
to move the body any effect upon the work done ? 

460. A man weighing 140 pounds carries a load of 50 pounds 
up a ladder 40 feet long inclined at an angle of 60° with the 
horizontal; how much work does he do? 

461. How much work is required to up-end a telephone pole 
35 feet long and 1 foot in diameter, if it weighs 50 pounds per 
cubic foot? 

462. A circular well 4 feet in diameter is dug 20 feet deep. 
Find the work done in raising the material to the surface if the 
material weighs 140 pounds per cubic foot. 

463. Find the work done in raising to the surface the water 
from a full cistern which is 10 feet square and 6 feet deep. 

464. Find how many units of work are stored up in a mill 
pond 100 feet long, 50 feet broad, and 3 feet deep, the point 
where the water is discharged being 11 feet below the surface of 
the pond. 

465. An iron plate -| inch thick has the form of an isosceles 
trapezoid whose altitude is 6 feet, and whose bases are 15 feet 
and 5 feet. This iron plate rests on a level pavement. The 
shorter base is lifted until it is 4 feet above the pavement, the 
longer base remaining where it was. Find the work done in 
foot-pounds if the iron weighs 450 pounds per cubic foot. 

137. If a constant force F, acting on a particle, displaces it 
a distance s along a line which makes an angle with the line 
of action of the force, the work done on the particle is defined 
by the equation 

Work = i^cos^X5; 



Work and Energy 



235 



that is, the work is the product of the component of the force, 
in the direction of motion, and the displacement of the particle. 

If the component of the force in the direction of motion is 
called the effective force, the work done is the product of the 
effective force and the displacement of the particle. The work 
IS positive when the effective force is positive, and is negative when 
the effective force is negative. 

The work done on a particle by the simultaneous action of 
any number of forces, constant in magnitude and direction, is 
the work done on it by the component, along the line of motion 
of the particle, of the resultant of all the forces which act on it. 

Example 1. — A block weighing 100 pounds is dragged, at a 
uniform speed, 100 feet along a rough horizontal plane by a 




V 

100 
Fig. 202. 



force of 20 pounds, whose line of action makes an angle of 60^ 
with the plane. Find the work done on the block by the force. 
Solution.— {'^QQ Fig. 202.) 

Effective force = 20 x cos 60° = 10 lbs. 
Therefore, work = effective force X displacement 
= 1000 ft.-lbs. 

Example 2. — A body weighing 67 pounds is drawn, at a uni- 
form speed, 85 feet up a rough plane inclined at arctan y\ to the 



236 



Elementary Mechanics 



horizontal, the drawing force being horizontal. If the coefficient 
of friction is J, find the work done : 

(a) by the force on the body, 

(b) against the force of gravity, 

(c) against the force of friction. 

Solution. — By Newton's second law of motion, the resultant 

N 




Fig. 203. 



force lip the plane is zero. Therefore, resolving forces perpen- 
dicnUir to and along the plane (Fig. 203), we lind 

i^sin^ = iY-67cos^, (1) 

and 

i^cos^ = iV/5 + GTsiu^. (2) 

From (1) and (2) we get 

N = S5 lbs. and ?^ cos 6' = 48y\lbs. 

(a) AVork = effective force X distance = -J:8yVx 85 = 41 2 5 ft.-lbs. 

( b ) AVork = effective force X distance = 

(G7Xtt) X85r=2(;80ft.-lbs. 

(c) Work = effective force X distance = -«/ x 85 = 1445 f t.-lbs. 

Prohlems 

466. A weight of 20 pounds is dragged 50 feet up a plane 
inclined at 30° to the horizontal by a constant force P = 25 
pounds, acting at an angle of 15° to the plane. There is a 
retarding force R — 5 pounds along tlic plane. Compute the 
work done by or against each force. 



Work and Exergy 237 

467. The frictional resistance along an incline of 1 in 25 
being 150 ponnds per ton^ how much work is done when 2 tons 
are dragged at a nnif orm speed 100 feet along the plane ? 

468. The frictional resistance along an incline of 1 in 30 
being ISO pounds per ton, how much work is done when 3 tons 
are dragged at a uniform speed 90 feet along the plane? 

469. A weight of 26 pounds is moved 13 feet up a rough plane 
(^z=^) inclined at an angle of arctan -f^ to the horizon, by a 
force of 20 pounds acting at an angle of arctan f to the inclined 
plane. Find the work done against gravity, and against friction. 

470. A weight of 39 pounds is moved 13 feet up a rough plane, 
(^t = i)^ whose base is 12 feet and whose height is 5 feet, by a 
constant force of 50 pounds acting parallel to the plane. Find 
(a) the work done against friction, and (b) the work done 
against gravity. 

471. Find the work done if a body weighing w pounds is 
dragged up a rough plane whose length is I feet, base d feet, and 
height h feet. The coefficient of friction is /.i, and the velocity is 
constant. 

138. Energy. — By the energy of a particle is meant the work 
which the particle possessing it is capable of doing. It appears 
then that the unit of energy is the same as the unit of work. 

If a stone weighing 5 pounds falls freely through a distance of 
100 feet the force of gravity does 500 foot-pounds of work. It 
is thus seen that the stone, when at the elevation of 100 feet, 
has 500 foot-pounds of energy stored up in it. The energy which 
a particle possesses by virtue of its position of advantage is 
called potential energy. Thus a particle of weight TT at a 
height H has TT'x^ units of work stored up in it, or possesses 
^yxH units of potential energy. 

It is a matter of experience that every moving particle can do 
work upon another particle. It follows then that every moving 
particle possesses energy. The energy which a particle possesses 
by virtue of its motion is called kinetic energy. 



or 



238 Elementary Mechanics 

The total energy of a body is the sum of its kinetic and poten- 
tial energies. In most systems which we shall consider, this sum 
remains constant. Snch systems are called conservative. 

139. Expression for Kinetic Energy. — Suppose a particle of 
mass M, moving in a straight line with a velocity v, is brought 
to rest in a space s by the action of a constant force whose com- 
ponent along the line of motion is F. From Newton's second 
law of motion, 

-F=Ma = MJ/^ =M. ^ . '^^ =Mv. ^, 
at at CIS as 

the minus sign being used because the force F tends to decrease 
the velocity. It follows then that 

{' Fds=-\\Mvdv, 

Jo J V 

ds = M\ V dv. 

Therefore, 

Fs = iMvK 
But Fs is the work the particle can do in consequence of its 
motion. Hence the kinetic energy of a particle of mass M 
moving in a straight line with a velocity v is ^Mv^. For example, 
the kinetic energy of a body of weight 64.4 pounds moving with a 
velocity of 25 feet per second is 

Problems 

472. A body of weight 64.4 pounds has its velocity reduced from 
88 feet per second to 44 feet per second. Find the change in its 
kinetic energy. 

473. A body weighing 30 pounds slides down a rough inclined 
plane (iu, = j) of height 30 feet and base 100 feet. Find the 
kinetic energy acquired. 



Work and Energy 



239 



474. A body weighing 39 pounds is moved up a rough in- 
clined plane (/a — 0.25) of height 5 feet and base 12 feet, by a 
constant force of 50 pounds, acting parallel to the plane. Find 
the kinetic energy acquired by the body when it is 10 feet from 
the starting point. 

475. What is the kinetic energy of a car weighing 25 tons 
moving 6 miles per hour and loaded w^th 36 passengers of aver- 
age weight 154 pounds? 

476. A weight of TF pounds is drawn up a smooth plane, in- 
clined 30 degrees to the horizontal, by an equal weight which 
falls vertically: the weights are connected by a string which 
passes smoothly over the top of the plane. If the tirst weight 
is initially at rest and at a distance of 16 feet from the top of 
the plane, find the combined kinetic energy of the weights, just 
as the first one is about to slip off the plane. 

140. Work Done by a Force, Constant in Direction but Vari- 
able in Magnitude. — Suppose a variable force, acting on a 
particle, displaces it from Po to P (Fig. 204), along the line of 




Fig. 204. 



action of the force. Let P be the position of the particle at any 
time and denote by s the distance from Po to P. In this con- 
sideration F is assumed to be a function of s. AYe now seek a 
definition of the work to be done on the particle by the force. 

Let the interval from Pq to P be divided into n equal parts and 
designate by As the common length of these sub-intervals. De- 
note by Fi the value of F at the beginning of the ith interval. 
If we assume that Fi remains a constant throughout the ith 
interval, the work done on the particle by F would be 

P1A5 + P0AS + P3A5+ .... +PhAs. 



240 Elementary Mechanics 

The limit of this sum, when n increases without limit, is the 
integral of Fds between the limits and s, and we are led to define 
the work done in this case by the equation 



Work= Fds. 

Jo 
If s is measured from a point on PoP, such that So is the coor- 
dinate of Po and s the coordinate of P, then 

=r Fds. 



Worker 

J So 



In finding the work done in any particular case F is expressed as 
a function of s and then Fds is integrated between the proper 
limits. 

Example 1. — A weight of 400 pounds compresses an elastic 
spring 1 inch. Find the work done in compressing it 6 inches. 
Solution. — By Hooke's law 

F = ks. 
Since 5=1 when i^ = 400, 

yl' = 400 and i^=:^5 = 400s lbs. 
Therefore 



k=r Fds=\' 

J So Jc 



Work=l Fds=\\msds^l20^ in.-lbs. 



Example 2. — Find the work done on. a stone of weight 10 
pounds which falls from interstellar space to the earth's surface 
under the action of the force which the earth alone exerts. 

Solution. — Denote by P the radius of the earth and by s the 
distance of the stone from the earth's center at any instant. Then 

Since F=— 10 when s — P, 



-10 = A'/i?- and F^-\^P-/s''. 




Therefore 


Work= r Fds^ r - lOi^^ ^ = lOi^^ 

Joo J 00 ^ J 


r ds 

R S^ 


= 10i2 = 211,200,000 ft.-lbs. 





Work axd Exergy 



241 



Example 3. — An engine which uses compressed air has a 
C3'linder of length 24 inches and a cross-sectional area of 16 
square inches. Air at a pressure of 100 pounds per square inch 
is admitted into the C3dinder during the first half of the stroke. 
The valve is then closed. Find the work done on the piston 
during the forward stroke, assuming the expansion of the air 
obeys the law 



yv 



a constant, 



p being the intensity of pressure and v the volume of the air at 
any instant. 

Solution. — (See Fig. 205.) The working force for the first 



l^ 



Fig. 205. 



half of the stroke is 

-Fi =: pressure X area = 100 X 16 pounds, a constant force. 

The working force for the second half of the stroke is 

i*^o = pressure X area =rpx 16 pounds, a variable force. 

Since p = 100 when t' = 12 x 16, we get 

^• = 100(12xl6)l•'^«5 
and 



Ic 



1 •■^. V 1 f;\ l-^OS /I 0\ 1.408 

= 100 ( -^-x^^> ] =100^^^ 



p 
Therefore the working force for the second half of the stroke is 



16 



/'\9\ 1.408 

F, = pxl6 = 1600 l^j lbs. 



242 Elementary Mechanics 

The work done is then 

(24 r24 j^ 

i^o6^a:=12(1600)+1600(12)i-4°H -^, 
12 J 12 •^' 

^^^2QQ^ 1600x12 |-i_(o.5)«-'^°«], 

. 408 
=: 30790 m.-lbs. = 2566 ft. -lbs. 

141. If the force F makes an angle 6 with the direction of 
motion the work is defined by the equation 

Work=l F cos Ods. 



i: 



That is, the work is equal to that done by the effective force. 

In case a number of forces, constant in direction and variable 
in magnitude, act on the particle simultaneously, the work done 
by them on the particle is the same as that done by the com- 
ponent, along the line of motion of the particle, of the resultant 
of all the forces which act on it. 

Prohlems 

477. A weight of 300 pounds hangs by a chain 100 feet long 
which weighs 2.5 pounds per foot. How much work is required 
to lift the weight by winding up the chain on a windlass? 

478. A spring is compressed from 18 inches to 14 inches by a 
weight of 504 pounds. How much work must be done to com- 
press it two more inches? 

479. A weight of 500 pounds hangs by a chain 500 feet long 
which weighs 2 pounds per foot. How much work is necessary 
to lift the weight by winding up the chain on a windlass ? 

480. It requires a force of 960 pounds to compress the spring 
of a safety valve from 20 inches to 17 inches. How many foot- 
pounds of work must be done to compress it 1 inch further? 

481. A steam engine has a cylinder 18 inches long with a 
diameter of 12 inches. Steam with boiler pressure of 120 pounds 
per square inch is admitted into the cylinder during the first 
third of the stroke. The port is then closed, and assuming that 



Work axd Eneegy 243 

the pressure due to the expansion of the steam follows Boyle^s 
law {pv = a constant), find the work done on the piston during the 
forward stroke. 

482. A spring is normally 21 inches long; a force of 5 pounds 
compresses it 4 inches; how much work would be done in com- 
pressing it so that its final length is 12 inches ? 

142. Power. — A force, however small, can do a work of any 
magnitude provided sufficient time is allowed. The work re- 
quired to lift 100 tons of coal from a mine 180 feet deep is the 
same whether the task is performed in one hour or in eight hours. 
Since in practice time is a most important factor, it is necessary, 
in comparing agents which do work, to consider the time em- 
ployed. The rate at which work is done is called power. Hence 
the unit of power is the unit of work divided by the unit of 
time. The horsepower (H. P.) is the unit of power commonly 
used in engineering practice, and may be defined as the power of 
an agent which can do 550 foot-pounds of work per second, or 
33,000 foot-pounds of work per minute. For example, to hoist 
3300 pounds of mortar to the top of a structure 150 feet high in 
5 minutes would require 99,000 foot-pounds of work per minute, 
or 1650 foot-pounds of work per second; therefore an agent which 
can deliver 3 H. P. is needed. 

Example 1. — What horsepower is necessary to keep a 150-ton 
train moving on a level track at 40 miles per hour, the frictional 
resistance being 20 pounds per ton ? 

Solution. — Since there is no acceleration, the force exerted on 
the train must equal the total resistance. 

i^ = 20xl50 = 3000 lbs. 
By definition, 

TT-p_ Fs _ Fv _ 3000 X 176/3 _ ^^^ 

^'^- bbOTt - 550 - 550 -^^"• 

Example 2. — A steel rod 12 square inches in cross-sectional 

area and 6 feet long is to be stretched 1100 times per minute. 

Find the horsepower required to produce in it a tension of 72,000 

pounds. (£' = 3 X 10' Ibs./in.-. See Arts. 85-87.) 



244 Elementary Mechanics 

Solution. — Work per stretch = average force X distance 
stretched ; 

y / 72000 \ 
w F ^, F Ip 72000 V 13 7 p,iQ 1 • 11 

^ p^ 518.4X1 100^ 

12x33000 • * 



Problems 

483. How many pounds per hour will a 10 PI. P. engine hoist 
from a depth of 600 feet, supposing 20 per cent of the power is 
lost in friction? 

484. What H. P. is required to keep a 200-ton train moving 
on a level track at 40 miles per hour, if the frictional resistance 
is 20 pounds per ton ? 

485. A 600-ton train is moving at 15 miles per hour up a grade 
of 1 in 200. If the frictional resistance is 10 pounds per ton, 
what horsepower is the engine developing? 

486. A motor car weighing 0.75 tons runs (without power) 
down a grade of 1 in 30 at 12 miles per hour (no acceleration). 
What horsepower must it have to run up the same grade at the 
same speed? 

487. What horsepower is necessary to keep a 100-ton train 
moving on a level track at 30 miles per hour, the frictional resis- 
tance being 20 pounds per ton? 

488. A train weighing 200 tons is moving up a slope of 1 in 
85 at 20 miles per hour. If the frictional resistances are 12 
pounds per ton, what horsepower is the engine developing? 

489. A train weighing 80 tons is climbing a 2 per cent grade, 
the frictional resistances being 10 pounds per ton. If the engine 
is developing 440 horsepower at the instant the speed is 30 
miles per hour, what is the acceleration at that instant ? 

490. An engine pulling a certain load exerts a constant force 
of 500 pounds and travels at the rate of 20 miles per hour, (a) 
How much work does it do in 5 minutes? (b) What horsepower 
is it developing? 



i 



Work and Energy 245 

491. A 6-inch rapid-fire gun cliscliarge 5 projectiles per 
minute, eacli weigliing 100 pounds, with a velocity of 2800 feet 
per second, (a) What kinetic energy does each projectile possess 
on leaving the gun? (b) What horsepower is the gun devel- 
oping ? 

143. Relation Between Kinetic Energy and Work. — Suppose 
a particle of mass M moves from Po to P-^ under the action of 
any number of forces, constant in direction, and denote by F 
the component, along the line of motion of the particle, of the 
resultant of all the forces which act on it. Let Vo be the velocity 
of the particle at Po and v-^ its velocity at P^. 

From Newton's Second Law of Motion 



Then 



at at CIS as 



r^ Fds= P' Mv dv = UIv^^-mvo' 

JPo JVo 



That is, the change in the kinetic energy of a particle equals the 
work done on it by the component, along the line of motion of 
the particle, of the resultant of all the forces which act on it. 

Example 1. — Find the work done on a body of weight 64.4 
pounds by a force that changes its velocity from 10 feet per second 
to 20 feet per second, in a distance of 40 feet against a uniform 
resistance of 25 pounds. 

Solution. — The force must overcome the resistance and also 
produce the change in velocity. 

The work required to overcome the resistance is 

Work = force x distance = 25 x 40 f t.-lbs. = 1000 f t.-lbs. 
The work required to produce the change in velocity is 

Work=:iM(V-^o')=:i ^ (400-100) ft.-lbs.r=300 ft.-lbs. 

The total work done by the force is then 

1000 ft.-lbs.-F- 300 ft. lbs. = 1300 ft.-lbs. 



246 



Elementary Mechanics 



Example 2. — A weight W is projected clown the rough incline 
AB (/A = 0.25) from B with an initial velocity of 10 feet per 



K 



^^3 




B 



lOOft. 



w 



Fig. 206. 



second. Find the distance W will travel on the rough horizontal 
plane AE {[x — l) before coming to rest. (Assume no shock at ^.) 
(See Fig. 206.) 
Solution. — The kinetic energy of TF at 5 is 

50 






32.2 



W. 



The kinetic energy acquired by W in sliding from B to A is 
the work done on the particle : 

W 

(F sin 30°--^ cos30°)>:200. 

Hence the kinetic energy of W at A is 

|5^ + (Fsin30°-5'cos30°) x20C= |^ +10017-25 V 3 17. 

The body will move along AE until its kinetic energy is over- 
come by the resistance lf/3. 

Hence, if 8 denotes the distance W travels along AE, we have: 



and 



^ XS= 11^ +100F-25V3F 



^=174.8 ft. 



Work and Energy 



247 



Problems 

492. A car weighing 26^^ tons breaks loose on a level track 
from a train moving 15 miles per hour. The frictional resistance 
(constant) is 300 pounds. How long before the car will stop, 
and how far will it have gone from where it broke loose ? 

493. A weight of 20 pounds is projected along a rough hori- 
zontal plane with an initial velocity of 8 feet per second. _ If 
the frictional resistance is 4 pounds, how far will the weight 
move before coming to rest? 

494. A 30-ton car is moving with a velocity of 30 miles per 
hour on a level track. The brakes refuse to work. How far 




Fig. 207. 

will the car go after the power is turned off before coming to 
rest if the friction is 1 per cent of the weight of the car ? 

495. An automobile going at the speed of 30 miles per hour 
comes to the foot of a hill. The power is then shut off. The hill 
rises 1 in 50. How far up the hill will the automobile go if the 
friction is 6 per cent of the weight of the car ? 

496. A bodv of weioiit 17 starts at A and slides down a rouo^h 
plane (ya = 0.25) 100 feet long and inclined at an angle of arctan 
0.75 to the horizontal. At the lowest point, B, of the plane the 
body runs on to a rough horizontal plane (/;i = 0.125). Find the 
distance from B the body comes to rest. (Assume no loss of 
kinetic energy at B.) (See Fig. 207.) 



248 Elementary Mechanics 

497. Solve last problem if the body has at A an initial velocity 
of 32.2 feet per second. 

Review ProhJems 

498. A vat which is a right prism whose cross section is an 
isosceles triangle is full of fresh water. The length of the vat is 
20 feet;, its depth is 9 feet, and the distance across its top. is 
12 feet. Find the work done in pumping the water out of the 
vat, if it is to be discharged out of an orifice 8 feet above the 
top of the vat. A cubic foot of fresh water weighs 62.5 pounds. 

499. Find the work done in emptying a right circular cone 
of height h and radius of base a of a liquid which weighs w 
pounds per cubic foot. The axis of the cone is vertical and its 
vertex is down. 

500. A rectangular tank, 3 feet long, 2 feet wide, and 1.5 feet 
deep is filled from a cylindrical tank of 24 square feet horizontal 
cross section, in which the water level at the start Is 20 feet below 
the bottom of the rectangular tank. How much work is done, 
assuming that the intake pipe enters the bottom of the rectangular 
tank ? 

501. An engine of 10 horsepower raises 4000 pounds of coal 
in an hour from a pit 1200 feet deep, and also gives motion to a 
hammer which makes 50 lifts of 5 feet each in a minute. Find 
the weight of the hammer, assuming the efficiency of the engine 
is unity. 

502. A chain weighs 10 pounds per foot and is 300 feet long. 
It hangs from a drum into a mine shaft: How much work is 
done in winding up 100 feet of it? 

503. A weight of 300 pounds hangs by a chain 100 feet long. 
The chain weighs 2.5 pounds per foot. How much work must be 
done to lift the weight 50 feet by winding the chain upon a 
windlass ? 

504. What horsepower is necessary to keep a 150-ton train 
moving ojq a level track at 40 miles per hour, the frictional 
resistance being 20 pounds per ton? 

505. A 2-ton car runs down a slope of 1 in 20 at 10 miles per 
hour, the resistance at that speed being just sufficient to prevent 
acceleration. What horsepower must it exert to run up the 
slope at the same speed ? 



Work and Enkrgy 249 

506. What is the effective horsepower of a locomotive which 
gives a 500-ton train a speed of 15 miles per hour up a grade of 
1 in 320, the frictional resistance being 12 pounds per ton ? 

507. A car weighing 3000 pounds runs down a slope of 1 in 20 
at 15 miles per hour, the resistance at that speed being just 
sufficient to prevent acceleration. What horsepower must be 
exerted to make the car run up the slope at the same speed ? 

508. A 400-ton train is moving up a grade of 1 in 100 against 
a frictional resistance of 10 pounds per ton. What horsepower will 
the engine develop at the instant when the speed is 15 miles per 
hour, if the acceleration of the train at that instant is 0.2 feet per 
second per second ? 

509. An 8-ton trolley car starts at A with a speed of 15 miles 
per hour, and reaches B with a speed of 22.5 miles per hour: if 




Fig. 208. 

the average frictional resistance is 15 pounds per ton, and the 
driving force remains constant, what horsepower is being devel- 
oped at 5 f (Fig. 208.) 

510. What is the horsepower of an engine that can raise every 
minute and a half 500 cubic feet of water to a height of .100 feet? 

511. Find the horsepower of an engine which is drawing 120 
tons up an incline of 1 in 300 at 30 miles per hour against a 
frictional resistance of 20 pounds per ton. 

512. A slider weighing 100 pounds rests on a table. It is 
moved by a weight of 20 pounds fastened to it by a rope which 
passes over a pulley at the edge of the table. When the slider has 
moved 2 feet its velocity is 2 feet per second. Find the coefficient 
of friction. 

513. In the motion of a particle down a smooth cycloid the 
vertical component of its velocity is greatest when it has com- 
pleted what fraction of its vertical descent ? 



250 



Elementary Mechanics 



514. The average pressure on the piston of a steam engine is 
60 pounds per square inch, the area of the piston is 1 square foot, 
the length of the stroke is 18 inches, the pressure occurs only 
in the forward motion. How many strokes does the engine make 
per minute if it registers 8 horsepower? 

515. Unit particles, which attract each other according to the 
law of the inverse square, are situated at 
A and B. The particle at B is held fast 
and the particle at A is constrained to 
move in the arc of the circle. Find the 
amount of work done by the attracting 
force when the particle at A has moved 
one-third the way to B. (Fig. 209.) 

516. A particle of mass M is fixed at the 

origin and attracts according to the law of 

the inverse square a particle of mass m 

which is constrained to move in the line 

y = 3. Show that the work done by the attractive force in moving 

the particle from the point where x = 4ca to the ^-axis is 4:Mm/5a. 

517. Find by integration the work done against gravity in 

bringing a weight w from Ato B along the circle. (See Fig. 209.) 




Fig. 209. 



CuRviLiNEAK Motion 251 

CHAPTER X 

CURVILINEAR MOTION 

144. Vector Velocity. — When we considered motion along a 
straight line, the location of a moving particle at any instant 
could be given by a scalar quantity, namely, its distance from a 
fixed point on the line. We shall now consider the motion of a 
particle in a curve. To determine its position we must now have 
given a vector, namely, that from some fixed point as origin to 
the position of the particle. This is called the position vector of 
the moving particle. As the particle moves, this vector changes in 




Fig. 210. 

direction or magnitude or both. Let p and p + Ap (Fig. 210) be 
the position vectors of a particle at the beginning and at the end of 
a time interval A^. Then the vector difference Ap (see Art. 6) will 
be as indicated in the figure. If this difference is divided by the 
scalar A^ (see Art. 4) we shall have a new vector Ap/At having the 
same direction as A/a but differing in magnitude. If now we let 
At approach zero, Ap will also approach zero, but Ap/At will in 
general approach a vector value differing from zero in magnitude 
and having a definite direction. This limiting vector is called the 



252 Elementaey Mechanics 

vector velocity of the particle at the beginning of the interval Al 
I^ may also be called the derivative of the position vector p with 
respect to the time. Its direction is along a tangent to the curve 
(from the definition of a tangent). Its magnitude is the speed 
of the particle. 

145. Vector Acceleration. — In any curvilinear motion the 
velocity vector is itself constantly changing, perhaps not in magni- 
tude, but certainly in direction. It will itself have a vector 




Fig. 211. 

derivative formed in the same way. Thus let a and cr + Ao- (Fig. 
211) be the velocity vectors at the beginning and at the end of a 
time interval Ai^. Form the vector difference Ao-, divide it by A^^ 
and let A^ approach zero. The limit of the quotient Aa/M is the 
vector acceleration of the moving particle. It is evident that it 
must point somewhere toward the inner or concave side of the 
curve. 

146. Composition and Resolution of Velocities and Accelera- 
tions. — If the curve we have considered lies in a plane, and we 
take coordinate axes in that plane with an origin at 0, the com- 
ponents of p in the directions of the axes are the coordinates, 
X and y, of the moving particle (see Fig. 212). It is also clear 



CuRviLiNEAE Motion 



25; 



that the components of Ap are Lx and Ly, those of Ap/At are 
Y 




Fig. 212. 

Ax/At and Ay/ At, and, finally, passing to the limit, the com- 
ponents of o- or dp/dt are dx/dt and dy/dt. These are called 
Vx and Vy, respectively, or the x and 
y components of the velocity (see Fig. 
213). If in Fig. 212 we replace p, x, y, 
A/0, Ax, and Ay, by a, v^, Vy, Aa, Avx, 
and Avy, respectively, we can show in 
like manner that the components of a or 
da/dt are 




dv 



o-x — 



^'^^'andaj,== -^ 



dvr 



d'-y 



Fig. 213. 



dt dt- 

The magnitude of a, i. e., the speed of 

the particle, is called v, and is equal to V^^j," + ^'/. It is shown in 
the Calculus that v — ds/dt, where s is the distance measured 
along the curve. Similarly the magnitude of the acceleration a is 



It is most important for the student to remember that in curvi- 
linear motion velocity and acceleration are both vectors, and that 
neither is determined by giving its magnitude alone. It is no 



254 



Elementaky Mechanics 



more correct to identify a velocity of 10 feet per second east with 
a velocity of 10 feet per second north than it would be to identify 
the former with a velocity of 20 feet per second east. 

Usually the most convenient way to specify the direction of a 
vector is to give the angle that it makes with a known direction^ 
such as the x-axis. Thus the angle that a makes with the x-axis is 

tan~^-l^ In determining the quadrant of the angle, however, 

we must take account of the signs both in the numerator and in 
the denominator separately, not merely of the sign of the fraction. 
Thus if the components of a vector a SiTe Vx= — \^S and Vy= +3, 
its angle with OX is 120°; but if i;^= + V3' and fy=-3, the 
angle is 300°. This matter will always be perfectly clear if we 
draw a figure. 

Example 1. — A point moves so that x — t- and ^ = log t. Find 
its velocity and acceleration when i^ = l. 
Solution. — Differentiating, 




Fig. 214. 



CUEVILINEAE MOTION 



255 



if we let a denote the magnitude of the acceleration. Here the 
velocity and acceleration are not equal although their magnitudes 
are the same. For their directions are different, o- pointing into 
the first quadrant, a into the fourth (see Fig. 214). If we 
denote by r and ^, respectively, the angles that the vectors o- and a 
make with OX, we shall have t = 26°34', and i/a = 333°26'. 

Example 2. — A particle P moves in the curve y- — \x so that Vx 
is always 2 ft/sec. Find v and a when P is 2 feet from the a;-axis. 




Solution. 



Hence, 



Fig. 215. 



2ydy = 4:dx. 

dii 2 dx 2v,. o 

dt y dt y 

^ = V2^+^"^=2V2"f/s. T = 45°; 



Hence, 



Vx- 

_4 



; . * . tto- = ; 



%■ 



4 dy 
y^'dt 



2. 



a = 2 f/s% 1/^ = 270°. (See Fig. 215.) 



Problems 

518. A particle moves in a curve y'^ — x^ so that ^^j/ = 4 ft /sec. 
Find V and a when ?/ = 8. 



356 Elementary Mechanics 

519. A point describes a curve 6y = x-, and has a speed of 
10 ft/sec when .t = 6. Find the components of the velocity at that 
instant. 

.520. A particle moves in the path y = 3x" so that the speed of 
its projection on OF is always 4 ft/sec. Find the direction of its 
motion 12 seconds after it passes the origin. 

x^ v^ 

521. A particle describes the ellipse ^ + "^r- = 1, so that 

Vx = S f/s. Find Vy and ay when y = 2 feet. Find also the time of 
describing one-half of the ellipse. 

522. Given Vx = 2y, Vy — 2x, show that the path is an equilateral 
hyperbola^ and that ax = ^x and ay — 4:y. 

523. If a particle moves so that .T = e* + e"* and y = e^ — e~\ 
show that the velocity and the acceleration have each the same 
magnitude as the position vector at any time t. Show also that 
r=:90°-i/.. 

147. Velocity and Acceleration Independent of the Origin. — 

In defining vector velocity we considered a fixed point as origin 
of the position vectors p and p + Ap. This was merely to preserve 
the analogy with the ordinary procedure of the Calculus. The 
vector Ap, however, is the vector from the position of the particle 
at the beginning of the time A^ to its position at the end; and 
this is certainly the same regardless of the origin. As the origin 
was not used again in forming a and ' a, these vectors are 
entirely independent of the origin, and depend only on the 
motion itself. At any stage of an investigation, therefore, we 
are at liberty to assume an origin, and if desirable a set of 
coordinate axes, in any w^ay that we please. If we desire the 
components of a vector in two given perpendicular directions, we 
may take coordinate axes in these two directions. We have seen 
that the velocity vector is always tangent to the curve. The 
acceleration vector, however, is not; and one of the most im- 
portant problems is to determine its components along a tangent 
and a normal to the curve. 



Curvilinear Motion 



257 



148. Tangential and Normal Acceleration. — To find at and a„, 
the tangential and normal components of the acceleration at any 
point, take the origin at that point and the a;-axis tangent to the 
path-cnrve (Fig. 216). Then at this point 

,, _ dy _ dy dx _Q _^ 
dt dx dt 

and therefore v — Vx at this point. In 
general 

Differentiating with respect to t. 



2v 



dt 



dt 



+ ^'^ dt 




But, since v^Vx and Vij — Q, we have 

dv _ dv^. _ _ __d-s Fig. 216. 

~di-~di -'''-''' -W 
Thus the tangential acceleration is the rate of cliange of the speed 
in the path. It does not depend on the shape of the path. It is 
zero if the speed is constant. 

To get the normal acceleration, we start with 

^ ^dy^dydx^ 
^ dt dx dt * 
Differentiating, 



ttn — Cly' 



dv, 
dt 



dy d-x 
dx 

1 



dt~ dx 



m 



= 0.a..+ 



since the expression for curvature, 1/r, developed in the Calculus, 

reduces to -—^ when -,- =0. The reason that we make this sub- 
dx- dx 

stitution is in order to get a result independent of our choice 
of axes. 



17 



258 Elementary Mechanics 

We have then the very important expression for the normal 
acceleration. 



where r is the radius of curvature at the point considered. Thus 
we see that the normal acceleration depends on the speed and the 
shape of the path, but not on the rate at which the speed is 
changing. 

It will be seen that the expressions that we have just derived 
for at and an are not expressed in terms of coordinates. They 
are therefore valid regardless of the coordinates used. The 
student should try to visualize the acceleration vector, observing 
that it always points toward the inner side qf the curve, that it 
points along the normal to the curve if the speed is constant, 
ahead of the normal if the speed is increasing, behind the normal 
if the speed is decreasing. 

'Example. — A point moves so that x = 2t^-]-3, y = t- + 2t. Find 
the tangential and normal accelerations when t=^2. 
Solution. — 

x = 2r- + 3, y = t- + 2t, 

Vx^^t, Vy=2t-j-2, 

ax = 4:, ay = 2. 

v^- = v/ + Vy' = 20r' + St + -^, , 



dv 
"^- dt 



20/ + -1-- 



V 



= 4.4. Ans. 

t = 2 



az^V 4^ + 2^ = 2V5; 
Also, a=Vai^ + a„". 

Hence, a,^ = Va--ar'= V20-19.36 = 0.8. Ans. 

Problems 

534. A particle moves so that .t = cos 2^ + sin 2t, ?/ = cos 2^ — 
sin 2t. Find the tangential and normal acceleration when i = 7r/4. 

525. A point moves in the curve y~ = 4:X so that Vx = 2i/s. 
Find the components of the acceleration when the point is 2 feet 
above the x-axis. 



CuKviLiNEAR Motion 259 

149. Circular Motion. — If a particle moves in a circle, its 
position at any instant is determined by the angle, 6, from a 
fixed line through the center of the circle, to a line joining the 

center to the particle. The rate of change in 9, — ^ is called the 

angular velocity and is commonly denoted by w. If (ji varies, its 

rate of chanoe is -^t or -^—- . This is called the anovular accelera- 
° at at- ° 

tion and will be denoted by /?. The four variables t, 6, to, and /5, 
bear a perfect analogy to t, s, v, and a, respectively, in rectilinear 
motion. The formulas connecting the elements of the latter set 
may be converted into the formulas connecting the elements of 
the former, Tvithout any further proof, but always subject to 
the corresponding restrictions. Thus in rectilinear motion, 
t'" = t'o" + 2a5 if the acceleration is constant. So in angular mo- 
tion, (x)- = ioq- + 2/36 (coq being the initial angular velocity) if the 
angular acceleration is constant. 

Prohlem 525a. — By the principle of correspondence just ex- 
plained, write the following formulas for angular motion: (1) 
connecting time, angle, and angular velocity, when the last is 
constant; (2) time, angle, initial angular velocity, and angular 
acceleration, when the last is constant; (3) t, Oj mq, and w, when 
13 is constant; (4) w, Wf^, ^, and t^ when (3 is constant; (5) ^^ Wq, 
to, and j3, when (3 is variable. 

150. Units. — In applying these formulas there is no restriction 
on the units employed except that they should be consistent. 
Thus if t is in hours and 6 in revolutions, to will be in revolutions 
per hour; or if m is in radians per second, 6 must be in radians 
and t in seconds. If the data are partly in one unit and partly 
in another, they must in general be made uniform before applying 
the formulas. 

Example 1. — A body changes its angular velocity from 
4 R. P. M. {revolutions per minute) to 40 E. P. M. in an interval 



260 Elementary Mechanics 

of 3 minutes. What is the angular acceleration (supposed uni- 
form), and how many revolutions will the body make in the next 
two minutes ? 
Solution. — 

(i= — ~^ — — ^ — = 12 rev/mm-. 
t o 

For the next 2 minutes 

^ = a)oi( + i^^- = 40x2 + 6x4 = 104 revolutions. 

ProMems 

526. A fly-wheel starting from rest acquires a speed of 
180 E. P. M. in making 7^ complete revolutions. Find the time 
and the angular acceleration. 

527. A motor is at rest and is brought to a velocity of 300 
radians per minute in ^ minute. Find the angular acceleration 
necessary and the number of revolutions made. 

528. A '^ merry-go-round " making 100 E. P. M. is brought to 
rest in 2 minutes. Find the angular acceleration and the angle 
turned through before coming to rest. 

529. A wheel is running at a uniform speed of 32 turns a 
second when a resistance begins to retard its motion uniformly at 
a rate of 8 radians per second per second; (a) how many turns 
will it make before stopping? (b) in what time is it brought to 
rest ? 

151. Linear and Angular Displacement, Velocity, and Acceler- 
ation. — If a particle moves in a circular path, the distance that 

it moves is determined if we know the 
angle turned through and the radius of 
the circle. If 6 is the angle in radians 
(Fig. 217), r the radius, and s the 
length of the arc passed over, then 

s = re. 

Since r is a constant, differentiating 
this twice with respect to t gives 

^^^- ^^'^- i; = faj and at — r/S, 




CURVILINEAE MoTION 261 

In both of 'these, also, it must be remembered that the angular 
velocity and acceleration must be expressed in terms of radians. 
The student should contrast this restriction with the freedom of 
choice of units in the preceding article, and remember that 
radians must be used if the angle is to be multiplied by the 
radius. Also it should be remembered that r/? gives the tangential 
and not the total acceleration. The other component, the normal 
acceleration, being equal to v~/r, may here be written w-r. 

Example 1. — A particle moves on the circumference of a circle 
with the constant angular velocity w. If the radius of the circle 
is r, find the magnitude and direction of the velocity and of the 
acceleration. 

Solution. — From the above the magnitude of the velocity is roj. 
Its direction is of course tangent to the circle. We have just seen 
that the normal acceleration is equal to w^r. Since w is constant, 
^ = and the tangential acceleration is zero. Hence the total 
acceleration is toV and is directed along the normal, i. e., toward 
the center of the circle. 

Example 2. — Solve the preceding example by expressing x and 
y in terms of t, and finding the velocity and acceleration from 
their components parallel to the axes. 

Solution. — Since the angular velocity is constant, 6 = o}t^ if we 
take ^ = when = 0. We then have 

x = r cos 6 = r cos o)t. y = f' sin ^ = r sin o)t. 

Vx= —ru) sin (Jit. Vy = rM cos wt. 



ax- 



r(ji"COSo)t. ay= —ro)^ sin (Dt. 



v = Vvx^ + Vy^ = ro). 



y 

2 



tanT= ^'^'^cQs^^ =-coto)/; T = a)]^ + 90° = ^ + 90° ; 

— rw sin ixit 

, , —r(x)- sin (lit , -i 1. n 
tan \p= ■ =tan (oc = tan 6. 

— ro)^ cos (x)t 

But each component of a is opposite in sign to the corresponding 
coordinate. Consequently if/ = 6-\-180° ; i. e., the acceleration 
vector points toward the center of the circle. These results agree 
of course with the preceding solution. 



262 Elementaey Mechanics 

Problems 

530. For motion in a circle, x = a cos I't, y = a sin ht, find 

ax} o.y, at, and an. 

531. If a point moves on an ellipse x — 4: cos t, y = 3 sin t, show 
that at/an= — 2-j sin 2^. 

532. A point moves in a circle of radius 10 feet with a uniform 
speed of 8 f/s. Find Vx and Vy when ^ = 30°. 

533. A point moves in a circle of radius 5 feet with a uniform 
speed of 6 f/s. Find the linear acceleration. 

534. A 14-foot fly-wheel is making 80 E. P. M. Find the 
linear velocity and the linear acceleration of a point on the rim. 

535. Taking the radius of the earth as 4000 miles, and 7r = 22/7, 
find the linear velocity and linear acceleration of a point on the 
earth's equator. 

536. A Ferris wheel of 15 feet radius, starting from rest, makes 
1/10 of a revolution during the first second of its motion. Find 
the angular acceleration of the wheel (supposed constant) . Find 
also the linear velocity and the total linear acceleration of a point 
on the rim at 2-| seconds from rest. 

537. The velocity of a particle moving in a circular path whose 
radius is 3 feet, changes from 7 revolutions to 4 revolutions per 
second in 5 seconds. If the acceleration is uniform: 

(a) What is the angular acceleration? 

(b) How many revolutions will the particle make before com- 
ing to rest ? 

(c) What was the normal acceleration at the beginning? 

538. If a particle moves in a circle with uinform angular 
velocity about the center, show that its angular velocity about any 
fixed point in the circumference is also uniform, and is equal to 
one-half of that about the center. 

152. Newton's Laws and the Differential Equation of Motion. — 

Up to this point in the study of curvilinear motion, we have been 
concerned only with Icinematics, i. e., the discussion of the motion 
itself without reference to the forces producing it. For the 
dynamics of curvilinear motion, i. e., the stud}? of the motion in 
relation to the forces producing it, we must generalize the state- 



Curvilinear Motion 263 

nient of Xewton's Laws and the differential equation of motion. 
(See Arts. 115, 118.) 

In curvilinear motion the First Law has obviously no applica- 
tion ; and the Third Law presents no new difficulty. The Second 
Law must be restated, and this can best be done in terms of 
vectors. We have seen that velocity is a vector quantity. Mo- 
mentum is also a vector quantity, being obtained by multiplying 
the velocity vector by the scalar factor of mass. The same will 
be true, of course, about change of momentum. We have also 
seen that force is a vector quantity. The Second Law may now be 
stated in the following form : the tivo vector quantities, force and 
change of momentum, are pro'portional, i. e., one is a scalar con- 
stant times the other. The statement in terms of vectors includes 
without explicit mention the latter part of the law, which states 
that the change of momentum is in the same direction that the 
force acts. Of course we must make the same convention about 
constant force and unit time that we did in rectilinear motion. 
(Art. 118.) These restrictions are removed when we pass to the 
differential equation of motion. This equation, \F = ma, holds 
without change, if we merely understand that F and a are vectors. 
Of course m and X are still scalars, the latter being simply a 
constant depending on the units used. 

153. Components of Force and Acceleration. — Since the force 

vector and the acceleration vector have the same direction, which 

makes, let us say, an angle 6 with the a;-axis, the resolved part of 

F in the x direction will be F cos 9, and that of a will be a cos 6. 

Denoting these components by X and ax respectively, we shall get 

d'^x 
XX = \F cosO = ma cos 6 = max — m • -j-~ . 

This is just the same equation of motion that we should get if the 
force Z acted alone. As the rc-axis may be chosen at will, this 
gives us the very important result that the resolved part in any 



264 Elementaky Mechanics 

direction of the total force acting on a body produces the same 
acceleration in that direction as if that resolved part were the 
only force acting. 

This principle furnishes a practical method for solving a great 
many problems on motion in a plane. A set of rectangular axes 
is chosen, usually horizontal and vertical, and the differential 
equation of motion in the direction of each of these axes is 
integrated separately. Of course the initial velocity is a vector 
and must be resolved into its horizontal and vertical components 
to get the initial velocities in these two directions. 

154. Free Motion and Constrained Motion. — There are two 
principal types of problem in curvilinear motion in a plane. In 
the first, that of free motion, the total force acting on the particle 
is known, and one part of the problem is to find the path which 
the particle describes. In the second, that of constrained motion, 
the particle is compelled to move along a certain path, which is 
known in advance, and one part of the problem is to find the force 
which must act on the particle to keep it in this path. Thus if a 
ball is thrown into the air (not vertically) we know the forces 
acting on it, namely gravity and air-resistance; but we do not 
know before solving the problem what the path is. This is an 
illustration of free motion. If, on the other hand, the ball is 
projected into a fixed tube, we know that its path must follow 
this tube; but we do not know before solving the problem what 
force the wall of the tube must exert on the ball in order to keep 
it in this path. This is an example of constrained motion. Other 
familiar examples of constrained motion are the sling-shot, the 
governor, a train moving on a curved track, etc. 

In free motion we generally know ax and ay. Integrating these 
and determining the constants by means of the initial conditions, 
we get X and y in terms of t. The elimination of t between these 
two equations gives us a single equation in x and y. This is the 
equation of the path curve. 



CURVILINEAK MOTION 



265 



155. Motion of a Projectile. — As an illustration of free motion 
we shall now consider the motion of a particle projected obliquely 
upward from a given point 0, with a given initial velocity. Let 
the x?/-plane be the plane of motion (Fig. 218), OX and OY the 
horizontal and vertical axes; and let the particle be projected 
from with an initial velocity of 1000 f/s directed along a line 




Fig. 218. 



which is inclined at an angle of 30° to OX. If we neglect the 
resistance of the air, gravity is the only force acting on the body 
during the flight. The acceleration in the horizontal direction is 
therefore zero, and that in the vertical direction is —g. Hence, 



a;r = -||-2=0, and a,;- ^ 



df' 



9- 



Integrating, 



dx _n 



dy 



= -9t + C, 



When ^ = 0, the component velocities are 1000 cos 30° and 1000 x 



266 Elementaey Mechanics 

sin 30°. Substituting these values in the last pair of equations, 

we have 

1000 cos 30° = Ci, 1000sin30° = + C2, 
and hence, 

v^^-^=500V3, Vy=-^=:-gt + 500. 
Integrating again, 

a::=500V'3^. ^=-£^+500/. (1) 

/J 

(The constants of integration are zero since x = y = when ^ = 0.) 
These equations expressing x and y in terms of t are the equa- 
tions of motion of the projectile. They determine completely its 
position at any instant. 

The student should notice that in these equations each element 
that determines the motion appears separately. The expression 
for X contains only the one term due to the horizontal component 
of the initial velocity. This is the same as for uniform motion, 
since there is no horizontal acceleration. The expression for y 
contains a similar term, 500^, which would appear alone if gravity 
did not act. But it also contains the term. — ^gt^, which would 
appear alone if the initial velocity were zero. Thus it is always 
easy to write the equations of motion for any projectile without 
going back to the velocity or acceleration, if we know the initial 
velocity. 

If we wish the equation of the path, we can eliminate t, and get 

gx^ 



X ---2 



!/ = 



V3 1,500,000' 
which is the equation of a parabola. Thus the path of the pro- 
jectile is parabolic. 

To find the point P where the projectile will strike the .T-axis 
in its descent {OP is called the range on the level), set y = in 
(1) and solve, getting first 

t= seconds, 



Curvilinear Motion 267 

which is the time of flight (taking g = 32). Substituting this 
value of t in the equation for x, we get 

a:=500V3 i^ = 27,100 ft., 

which is the range. 

To find the highest point of the trajectory, set Vy — 0. This 
gives 

t=—— =la.6 sees. 

which is the time it takes the projectile to reach the highest point 
in its trajectory. Putting this value of t in the equation for y, 
we get 



500 32 rsoo 

_32, 
which is the height of the trajectory. 



y = '''- 32 2 



= 3900 ft.. 



Problems 

539. A projectile is fired with an initial velocity of 320 feet 
per second in a direction making an angle of G0° with the hori- 
zontal. Find the range. 

540. Find the coordinates of the highest point reached by a 
body projected with a velocity of 256 f/s at an angle of 30° Avith 
the horizontal. 

541. Find the value of the angle which the initial-velocity 
vector should make with the horizontal in order that the range 
should be a maximum. {Hint: Express the range in terms of 
this angle and the initial velocity Vq, and then differentiate, 
regarding Vq as a constant). 

542. A projectile fired from the top of a tower at an angle of 
elevation of 45° strikes the level ground 60 feet from the foot of 
the tower at the end of 5 seconds. Find the initial velocity and 
the height of the tower. 

543. A body is projected with a velocity of 100 f/s at an angle 
of 75° with the horizontal. Find its range on a horizontal plane. 



268 Elementaey Mechanics 

544. A shot is fired at an angle of 45° with the horizontal and 
an initial velocity of 192 f/s. Find the greatest height and the 
time to reach that height. 

545. An airplane is moving horizontally with a velocity of 
96 f/s, when a body is dropped from the machine. Find the speed 
of this body 4 seconds after being dropped, and the angle between 
its motion and that of the airplane at that instant. 

546. What must be the initial velocity of a projectile to hit 
an object 3000 feet away and 408 feet high, if fired at an angle of 
40° with the horizontal? 

547. A man standing 15 feet from the foot of a pole 150 feet 
high, aims at the top of the pole. If the bullet just misses the 
top, where will it strike the ground if ^0 = 1000 f/s ? 

548. If a golf ball be driven from a tee horizontally with an 
initial speed of 300 f/s, where and when will it land on ground 
16 feet below the tee (neglecting air resistance). 

549. The velocity of sound is 1100 f/s. Find the range of a 
gun if the projectile and the sound of the discharge reach a pomt 
at the same instant. 

550. The big gun used in the World War projected its shell to 
a height of about 24 miles. If the distance from Laon to Paris 
is 76 miles, what was the muzzle velocity of the shell and the 
angle of elevation? (The student must for the present neglect 
air resistance ; but it should be remembered that the result is then 
only a very rough approximation.) 

551. A piece of ice is detached on a roof whose inclination is 
30°, at a point 8 feet from the eaves, which- are 24 feet above the 
ground. At what distance from the vertical plane through the 
eaves will it reach the ground ? 

156. Centripetal and Centrifugal Force. — We have seen that 
any particle moving in a curve has an acceleration toward the 
inner side of the curve. That is, the total acceleration in 
curvilinear motion always has a normal component, which we 
have called the normal acceleration. From Newton^s Second Law, 
or from the equation of motion, we know that there must be a 
force producing this, that is, the total force must have a normal 
component proportional to the magnitude of this acceleration and 



Curvilinear Motion 269 

to the mass of the moving particle. This normal component 
is called the centripetal force. If we are dealing with constrained 
motion, and this centripetal force is exerted by the constraint 
upon the moving particle, then there is an equal and opposite 
reaction exerted by the moving particle upon the constraint. 
This is called the centrifugal force. Thus if a stone is tied to a 
string and whirled around in a circle, the string exerts a force 
upon the stone, which continually deflects it from a straight-line 
path and compels it to move in a circle. This is the centripetal 
force. The reaction to this, namely, the force which the stone 
exerts upon the string, is the centrifugal force. 

We know (Art. 148) that the normal acceleration is v'^/r, 
where r is the radius of curvature of the path at the point con- 
sidered. If we take the equation of motion in the form gF—Wa 
(Art. 125), we have that the centripetal force is 

W W v^- 

r n— — an= — -, 

9 9 r 

directed toward the inner side of the curve. The centrifugal 
force, which is the one more often mentioned, is of course equal 
and opposite. 

If the path is circular, then r is simply the radius of the circle. 
In this case it will frequently be convenient to use orr instead of 
v^/r for the normal acceleration. 

Example. — An 8-pound weight is whirled in a circle on a 
smooth table at the end of a 4-foot cord which can just stand a 
tension of 9 pounds without breaking. Find the E. P. M. it is 
making at the instant that tho cord breaks. 

Solution. — 

.*. i; = 12 f/s, and (0 = 3 rad/sec, 
or 28.7 E. P. M. 



270 



Elementaky Mechanics 



Prolilems 

552. Find the centrifugal force of a weight of 64 pounds which 
is moving in a circle of 6-foot radius at a uniform speed so as to 
make one revolution in 3 seconds. 

553. A weight of 2 pounds is connected by a string 2 feet long 
to a fixed point of a smooth table. If the string can sustain a 
tension of 8 pounds, what is the greatest number of revolutions 
per second that may be made without breaking the string ? 

554. A flat horizontal plate revolves about a vertical axis, 
making 1 revolution per second. If |U,= 11/18, what is the greatest 
distance from the axis at which a weight can lie on it and not be 
thrown off by the centrifugal force ? 

555. Assuming g = 32 f/s^ and the radius of the earth 4000 
miles, in what time could a body revolve freely around the earth 
close to its surface without being pulled down by gravity ? 

157. The Conical Pendulum. — A string attached to a fixed 
point and supporting a weight W which moves in a horizontal 
circle with uniform speed, constitutes the simplest case of the 
conical pendulum. In order that the weight may actually move 
in a circle, there must clearly be some relation between its speed 
and the angle that the string makes with the vertical. Let it be 
required to find this relation, and also the tension in the string. 

There are actually only two forces 
acting on the body, its weight W, and 
the tension in the string T. (Fig. 
219.) The resultant of these must be 
just equal to the centripetal force 
necessary to produce the circular mo- 
tion. Now the centrifiigal force is 
equal and opposite to the centripetal, 
and hence equal and opposite to this 
resultant. Consequently it forms 
L^ with this resultant a system of forces 
in equilibrium — or, what amounts to 
^ the same thing, the three forces, T, 
W, and the centrifugal force, are in 
equilibrium. Calling the last Fn, 




Fig. 219. 



Curvilinear Motion 



271 



and applying the conditions of equilibrium by resolving hori- 
zontally and vertically, we have 

Tcos^=F; 
or, replacing Fn by arrW /g, 

W g 
Notice that the factor W cancels out and that tan is merely the 
quotient of the two accelerations, the centrifugal, (0^7% which is 
horizontal, and that of gravity, g, which is vertical. This is an 
easy way to remember this result, and it is equally easy to express 
tan in terms of the linear speed v instead of w, by using v-/r 
instead of wV. It is left as an exercise for the student to find the 
tension T. 

In the solution of this problem we have an example of a very 
important method by which a problem in kinetics is replaced by a 
problem in statics, by the device of introducing a force equal and 
opposite to that necessary to produce the motion. This force will 
then be in equilibrium with the forces actually operating ; and the 
problem is thus reduced to one in statics. 

Example 1. — The string of a conical pendulum is 12 feet long 
and makes an angle of 60° with the vertical. Find the tension in 
the string and the linear velocity of the bob, if the latter weighs 
5 pounds. (Fig. 220.) 

Solution. — Eesolving vertically, 
T.-| = 5, 
.*. T = 10 1bs. 

Eesolving horizontally, 

1 . 2 V ^— — • --7^ J 




32 6V3 

.'. i; = 24f/s. Fig. 220. 

Example 2. — A conical pendulum makes 38t-V R. P. M. 



The 



length of the string is 30 inches and the weight of the bob is 



272 



Elementary Mechanics 



4 pounds. Find the radins of the circle described, and also the 
tension in the string. (Take 77 = 22/7.) 

Solution. — 

w = — Vn = 4 rad/sec. 

Eesolving horizontally, 

rsin^=l^=2r. 
9 
Eesolving vertically, 

r cos (9 = 4, 

.*. tan 6= — . 

But from Fig. 221 




Fig. 221. 



Hence, 

/?= 2 ft. = 24 inches, 
r=18 inches, 

K/ 5- 

4- 



tan 0: 

r 

T 



Problems 

556. The cord of a conical pendulum is 3 feet long, and the 
bob, weighing 6 pounds, is making 40 R.P. M. Find the angle 
the cord makes with the vertical. 

557. A 40-pound weight swings at the end of a 6-foot string 
which is tied to a hook in the ceiling. How far below the ceiling 
is the weight when making 50 E. P. M. and what is the pull on 
the string? (Take 7r = 22/7.) 

558. A stone weighing 1 pound is whirled round by means of 
a string so as to describe a horizontal circle in a plane 2 feet 
below the point of suspension. Find the time of revolution and 
also the tension, the length of the string being 1 foot. 

559. A plummet is suspended from the roof of a railway car. 
How much will it be deflected from the vertical when the train is 
running 45 miles per hour over a curve of 300 yards radius? 



CUEVILINEAR MOTIOIST 273 

560. In the preceding problem what angle should the plane of 
the track make with the horizontal in order that the wheel pres- 
sure of the train should be normal to the track ? If the distance 
between the rails is 4 feet 8 inches, how much should the outer rail 
be elevated to give this angle. 

561. If d is the distance between the rails and li is the elevation 
of the outer rail, show that (&j^jproximately) 

gr 

if the pressure of the wheels is to be normal to the track. The 
quantities v, g, and r, have their usual meaning and should all be 
expressed in feet and seconds. The approximation consists in 
assuming the sine and tangent of a small angle equal to each other, 
the inclination of the track to the horizontal being so small in all 
ordinary cases that this is amply accurate for practical purposes. 

562. Show that in the preceding problem d may be taken in 
any convenient units (different from v, g, and r), and that then 
h will be given in the same units as d. 

563. A railroad curve has a radius of 1000 feet. Trains are to 
round the curve at 30 mi/hr. If the gauge is 4 feet 8 inches, how 
much must the outer rail be raised above the level of the inner in 
order that there shall be no lateral thrust on the rails ? 

564. A locomotive weighing 175 tons moves in an 800-foot 
curve with a speed of 40 mi/hr. Find the horizontal pressure on 
the rails if they are on the same level. 

158. Surface of a Rotating Liquid. — The problem of finding 
the shape of the surface of a liquid subjected to rotation is one 
arising in the study of hydromechanics. If an open vessel con- 
taining a liquid rotates with uniform velocity about a vertical 
axis through its center, the liquid will soon acquire an angular 
velocity equal to that of the vessel, as a result of friction. The 
free surface will assume a concave shape, symmetrical with re- 
spect to the axis of rotation; and we shall now show that this 
silrface is a paraboloid of revolution. 

Consider a vertical section of the vessel, OY being the axis of 
rotation, and x and ij the coordinates of a particle of weight W 
18 



274 



Elementary Mechanics 



in the free surface of the liquid. (Fig. 222.) Acting on this 
particle we have the weight W, which is vertical, and the centri- 
fugal force, which is horizontal. By Art. 74, the surface of the 

liquid must stand at right angles to 
the resultant R of these forces, so that 
R is perpendicular to the tangent as 
shown. 

For a given angular velocity, we 
have, then, 

^ . Wio^X 




Fig. 222. 





R cost = 


9 


Now 


.*. tanT = 
by Calculus 


^oj^X 

9 


tan T 


-dy 
- dx' 




' dx 


(i)^X 

~ 9 ' 




y= 


^9' 





and, integrating, 



the constant of integration being zero if we take the origin at the 
lowest point of the curve. Since this is the equation of a parabola, 
the surface of the liquid is a paraboloid of revolution. 

Example. — If a hemispherical bowl of radius 28 inches is filled 
with liquid and made to rotate about its vertical radius at 
45 E. P. M. how much will overflow ? 

Solution. — The paraboloid of revolution in this case must pass 
through the edge of the bowl; and the required volume is that 
of the portion of this paraboloid cut off by the plane of this edge. 
This volume is (by Calculus) equal to half the volume of the 
circumscribed cylinder. The radius of this is the radius of the 



Curvilinear Motion 



275 



bowl, 7/3 (in feet if we use g = 32). Its altitude is the value of y 
in the equation of the parabola when a: =7/3, 



y= 



^ ^ ^ a^ra)' ^ ^^ 



Hence, 



^9 



64 



256 



r = \.x^y = ix 5 • lU = 16.3 eu. ft. 



Frohlems 

565. A hollow paraboloid of revolution, altitude 2 feet and 
radius of base 8 inches, with axis vertical and vertex down, is half 
filled with liquid. What must its angular velocity about its axis 
be in order that the liquid may just rise to the rim ? 

566. Find the pressure on the top of a closed cylinder of radius 
r, just full of liquid of density iv, and rotating uniformly about 
its vertical axis with a velocity to. 

567. A hemispherical bowl of radius r, containing a given 
volume of water Y , is set rotating about a vertical radius. At 
what angular velocity does the water begin to overflow ? 

159. The Simple Pendulum. — A heavy particle suspended from 
a fixed point by a weightless, inextensible cord, and caused to 
move in an arc of a vertical circle under the action of gravity, is 
called a simple pendulum. The particle is called the pendulum 
hob. We may regard the forces acting on the bob as its weight ^ , 
the tension of the cord T , and the 
centrifugal force Fn due to the 
motion. The directions in which 
these forces act are indicated in 
the figure. 

Denote by A the fixed point of 
suspension, by I the length of the 
cord, by P the position of the 
particle at any time, by 6 the 
angle between the vertical AO and 
the cord AF , and by 5 the length 
of the arc OF. 




Fig. 223. 



276 Elementary Mechanics 

Suppose the pendulum bob is drawn aside from its vertical 
position through an angle a and then released. Since no motion 
takes place perpendicular to the path, we have at P 

T=W cos e + Fn. 

The only force which affects the motion is the component along 
the tangent to the path at P of the resultant of all the forces that 
act on the particle. This component is W sin 9. Since this 
force tends to decrease the distance s, the equation of motion of 
the pendulum bob is 

or, since s — W, 

-J- =:-^sin(9. 

This differential equation may be integrated by multiplying both 

members by 2 ,~ dt : 

-^ dt 

2'^^-f.dt:=^-2Umede. 

dt^ dt I 

This gives on integration 

Since , =0 when 6 = a, Cj^= — -^cos a, 
and therefore 

it follows, then, that 

y'^gi Veos^-cosa' 



CUEVILINEAE MOTION 277 

an integral which can not be expressed in terms of the functions 
used in elementary mathematics. 

However, when the angle of vibration of the pendulum is small, 
a close approximation to the motion is obtained by replacing sin 6 
by in the equation of motion. This then becomes 

which can be integrated completely. Since g and I are positive 
constants, this is the differential equation of simple harmonic 
motion. (Art. 133.) The time ^ of a single oscillation is then 

t=nVT7g. 

Thus the period is independent of the length of the arc through 
which the pendulum swings. 

160. The Seconds Pendulum. — A pendulum which makes one 
oscillation in one second is called a seconds pendulum. The 
length of such a pendulum is 

If the value of g is taken as 32.17, the length of the seconds 
pendulum is 

Z= — '-— feet = 39.12 inches. 

If the length I of a pendulum which beats seconds at a given place 
is known, the value of the acceleration of gravity at this place is 

g = 7rH. 

This is the method regularly used for determining the exact 
values of g at different points on the earth's surface. 

161. The Cycloidal Pendulum. — A particle of weight W sus- 
pended from a fixed point A by a weightless, inextensible, and 
flexible cord of length l^ and caused to move in a cycloid under 



278 



Elementaky Mechanics 




the action of gravity, is called a cycloidal pendulum. The forces 

acting on the particle are its 
weight W, the centrifugal force 
Fn due to its motion, and the 
tension T in the cord. The direc- 
tions in which these forces act are 
indicated in the figure. A cy- 
cloidal pendulum may be con- 
structed by causing the cord of 
the pendulum to wind on another 
O Wj, cycloid. 

^^^- 224. Denote by P the position of the 

particle at any instant, by 6 the inclination to the horizontal of 
the tangent to the path at P, and by s the length of the arc OP. 
Since no motion takes place perpendicular to the path, we have 

T=W cos e + Fn. 

The only force which affects the motion is W sin 6, a force 
which tends to decrease s. Therefore the equation of motion is 

or 

- = -gsme. 
The equations of the cycloid may be written in the form 

y = a(l-coscj,), 
and it can easily be shown by Calculus that a = 1/4: and (^:=^26. 
Therefore, 



ds=\/dx'-\-dif= ^V4(l + cos 2(9)2 + 4 sin^ 20dO = I cos OdO, 
and 



cos 6d6=l sin 6. 



Curvilinear Motion 279 

The equation of motion may now be written in the form 

d's __g 

which is the differential equation of simple harmonic motion. 
(Art. 133.) Therefore the time ^ of a single oscillation of a 
cycloidal pendulum is 

t=TrVl/g. 

Thus the period is independent of the arc through which the 
pendulum swings. It should be noted that this statement is exact 
for the cycloidal pendulum, as there was no approximation made 
in integrating its equation of motion. The corresponding state- 
ment made in the case of the simple pendulum was an approxima- 
tion, holding good only for small amplitudes. 



Problems 

568. Find the leno-th of a seconds pendulum where g = 32.20. 

569. Find the value of g at a place where the length of the 
seconds pendulum is 39.06 inches. 

570. How many oscillations per minute will a 40-inch pendu- 
lum make at a place where ^ = 32.16? 

571. If 39.11 inches is the length of the seconds pendulum at 
a certain place, how long must a pendulum be to make 10 oscilla- 
tions a minute at that place? 

572. A pendulum oscillates 50 times per minute at one place, 
and 50.3 times at another. Compare the values of g at the two 
places. 

162. Motion Along a Smooth Curve in a Vertical Plane. — Con- 
sider a particle of weight W starting from A with an initial 
velocity Vq and sliding down a smooth curve in a vertical plane, 
as shown in the figure. Denote by P the position of the particle 
at any instant, and by r the angle that the tangent to the curve 



280 



Elementaey Mechanics 



makes with the horizontal. The tangential force is then W sin r, 
and the equation of motion is therefore 




Since 


W 


sin T 


_W d's 
g dt^ ' 




sin T 


— - 


dy 
ds' 


-, d^s dv 
and-^TT; =v^ f 
dt~ ■ ds' 


we have 












W dv 
g ds 


ds 




Integrating, 







■ "' Since the particle started from A 

with a velocity v^, v = Vq when y = h, h being the original height, 

so that C=^ -i-gh. Therefore 

v^ = v^' + 2g(h-y). 

The distance h — y being the vertical projection of the path, this 
equation expresses the fact that the velocity acquired by a particle 
sliding down a smooth curve in a vertical plane is the same as if 
it had fallen freely through the corresponding vertical distance. 



Example 1. — A particle slides 
down a smooth vertical circular 
arc, starting 60° from the lowest 
point. Find the velocity at the 
lowest point if the radius of the 
arc is 6 inches. 

Solution. — 

/i = 5Crr-Jft; 
,'.i;=:4f/s. 




Fig. 226. 



CURVILIXEAE MOTIOX 281 

Example 2. — A weight is whirled in a vertical circle at the end 
of a cord 3 feet long, having just sufficient velocity to travel in 
the circle. What is the velocity at the highest and lowest points 
of the circle? 

Solution. — At the highest point the centrifugal force must be 
just enough to support the weight, or the centrifugal acceleration 
must be equal to the acceleration of gravity, or v-/r — g. Hence, 
vV3 = 32, or ?; = 4:V'6 f/s, at the highest point. At the lowest 
point the velocity will exceed this by the same amount as if it 
had fallen freely a distance equal to the diameter of the circle. 
Hence at the lowest point 

^2 :='?;o-+^^/i=: 96 + 2x32x6 = 480. 
.*. i' = 4V30f/s. 

Prohlems 

573. Find the velocity of the weight in the preceding example 
when on a level with the center of the circle. 

574. A pail of water is caused to swing in a vertical circle at 
the end of a cord 4 feet long. If the water as to remain in the 
pail find the necessary* velocity at the highest and lowest points. 

575. If the pail and the water in the preceding problem 
together weigh 6 pounds, find the tension in the cord when the 
pail is at a point 60° from the lowest point of the circle. 

576. A body is projected from the lowest point of a vertical 
circular track of radius 6 feet. If the body is prevented from 
leaving the track, what velocity at the lowest point will carry it 
around ? ^ 

577. By the method of this article derive an expression for 
the linear velocity of the bob of the simple pendulum at any point 
of its swing, and from this derive the formula for its angular 
velocity given in Art. 159. 

163. Work. — If any number of forces^ F^^F.^ . . . . , variable 
in magnitude and direction, act on a particle, which moves under 
their influence along some curved path, and if the (variable) 
angles which the lines of action of these forces make with the path 
are 6^, 0., .... , then the work done by the several forces will 
be respectively ^F-^ cos ^^ els, ^FoCOsOods . . . . , the integration 



282 Elementaey Mechanics 

extending over the whole path. If the resultant of all these forces 
is F, making an angle 6 with the path, then its resolved part, 
F cos 6, along the path will he equal at each point to the sum of 
the resolved parts, F^ cos B-^ . . . . , oi the components. Hence, 

\F QO^O ds—\F-^co^B^ds+]F^co%B^ds+ .... 
or 

The quantity of work done hy the resultant of any number of 
forces is equal to the algebraic sum of the quantities of work done 
hy the several forces during the same displacement. 

164. Kinetic Energy. — The total work done upon a body 
moving in any curved path may be found in terms of its initial 

and final velocities. For by Art. 153, F cos B=^m _,. Hence, 

a I 



F COS B ds = \ m--^~ds=\ 
So }So at jso 



ds dv J 
m--~ .—-ds = 
dt ds 



mv dv 



mv 



2 2 

Thus we see that in the most general case of plane curvilinear 
motion the work done upon a particle in changing its motion is 
equal to the increase in its kinetic energy, or the work done by the 
particle is equal to the decrease in its kinetic energy. 

This principle combined with the preceding, which enables us 
to Qonsider separately the work done by each force that acts, is 
particularly useful in considering constrained motion. Here 
the reaction of the constraint, which is the force that keeps the 
particle in its path, does no work ; for it acts at every point in a 
direction perpendicular to that of the motion. The change in 
kinetic energy, therefore, is equal to the work done by or against 
the remaining force or forces. 

Example 1. — The bob of a 5-foot pendulum weighs 2 pounds. 
If it starts from rest at its lowest point and is acted on by a 
constant horizontal force of 6 pounds, what velocity will it have 
when the cord is horizontal ? 



Curvilinear Motion 



283 



Solution. — In this motion work is done hy the horizontal force, 
and work is done against the force of gravity. The only other 
force acting is the tension in the cord, and this does no Avork. 
Each of the first two forces acts through a distance of 5 feet, the 
radius of the circle. Hence the work-energy equation becomes 



6x5-2x5 = 1. -.i;2. 
9 
Hence, 

i;=V20^=25.4 f/s. 

Example 2. — The bob of a simple pendulum of length 10 feet 
is drawn aside until it is 4 feet above the horizontal line through 
the lowest point of its path, and is then released. If the bob 
weighs 5 pounds, find the tension T in the cord when the bob is 
1 foot above the horizontal line. 

Solution. — The forces acting on the bob are as indicated in the 
figure. Eesolving forces along the 
cord, we get 



T = 5cos^ + 



To 



From the work-energy equation, 



5(4 



and 



''-io^'' 



0=), 



Therefore 



v'' = (3g. 



5(A) + ^ 



^=7.5 lbs. 




Problems 

578. What will be the angle with the horizontal made by the 
cord of the pendulum in Example 1 when the bob reaches its 
maximum elevation? {Hint: The kinetic energy will then be 
zero, and the work done can easily be expressed in terms of the 
angle.) 

579. Show that if a pendulum bob, starting from rest at its 
lowest point, is acted on by any constant horizontal force, it will 
rise to a point where the angle the cord makes with the vertical 



284 Elementaey Mechanics 

is twice as great as that when the bob is in equilibrium under 
the action of gravity and the horizontal force. 

580. If two equal bodies are projected with the same velocity 
at the two angles which give the same horizontal range, show that 
the sum of the kinetic energies at their highest points is inde- 
pendent of the angles of projection. 

581. A particle slides down the outside of a smooth vertical 
circle, starting from rest at the highest point of the circle. Find 
where it will meet the horizontal plane through the lowest point 
of the circle. 

582. A particle starting from a point (a, h) moves so that 
Vx and Vy vary as the corresponding coordinates; find the equa- 
tion of the path and the accelerations along the axes. 

583. In uniform circular motion show that the angular velocity 
about any point in the circumference is also uniform, and equal 
to one-half of what it is about the center. 

584. A particle describes the hyperbola x~/9 — y-/lQ = ly find 
(a) ax if Vy = 8, (b) ay if 7^^ = 12. 

585. A particle describes a parabola with such a varying 
velocity that its projection on a line perpendicular to the axis of 
the parabola is a constant Ic. Find the velocity and acceleration 
parallel to the axis of the parabola. 

586. Show that the velocity vector of a point describing a 
cycloid passes through the highest point of the generating circle. 

587. A chandelier weighing 80 pounds is suspended from the 
ceiling of a hall by means of a chain 12 feet long whose weight 
is neglected. By how much is the tension of the chain increased 
if it be set swinging so that the velocity at the lowest point is 
6f/s? 

588. A stone slides 12 feet without friction down a roof of 
inclination 30°. If the lower edge of the roof be 50 feet above 
the ground, (a) when, (b) where, and (e) with what velocity 
will the stone strike the ground ? 

589. A stone is dropped from a balloon, which, at a height of 
625 feet, is carried along by a horizontal air current at the rate 
of 15 m/h. Where, when, and with what velocity will it reach 
the ground ? 

590. A particle P moves in a curve y'^ — x^ so that Vy — 4: f/s. 
Find at and an when y — S. 



Curvilinear Motion 285 

591. Show that a projectile whose elevation is 60° rises three 
times as high as when its elevation is 30°, the magnitude of the 
initial velocity being the same in each case. 

592. A wheel 6 feet in diameter is making 50 E. P. M. when 
thrown out of gear. If it comes to rest in 4 minutes, find (a) the 
angular retardation, (b) the linear velocity of a point on the rim 
at the beginning of the retarded motion, (c) the same after 
2 minutes. 

593. A projectile fired from the top of a tower at an angle of 
elevation of 45° strikes the ground 60 feet from the foot of the 
tower at the end of 4 seconds. Find the height of the tower. 

594. A stone thrown with a velocity of 64 f /s is to hit an object 
on top of a wall 19 feet high and 48 feet distant. Determine the 
direction of the initial velocity. 

595. Find expressions for v,- and ve where r and 6 are the 
polar coordinates of a point. 

596. A spring is compressed from 20 to 17 inches by a weight 
of 471 pounds. How much work must be done to compress it 
another inch ? 

597. A body whose weight is 64.4 pounds falls freely from rest 
from a height of 5 feet upon a 200-pound helical spring. Find 
the compression in the spring. (A 200-pound spring is one that 
is compressed 1 inch by a force of 200 pounds.) 

598. Prove that the statement, " a rifle bullet does not rise 
more than one inch in a range of 100 yards," implies that the 
initial velocity must be greater than 2078 f/s. 



286 Elementaky Mechanics 

CHAPTEK XI 
DYNAMICS OF A RIGID BODY 

165. Rigid Body. — As stated in Chapter III we shall regard 
a rigid body as a collection of particles so connected that the 
distance between any two of them is constant. If the body is 
continuous we may regard it as made up of an infinite number 
of elements of mass dm and we may perform any summation 
extending over all the particles by integration. 

166. Plane Motion. — We shall confine our attention to motion 
in which all the particles of the body move along curves parallel 
to one fixed plane. This is called plane motion. We may think 
of the body as made up of lamina or thin flat slices parallel to this 
plane. These laminae are all rigidly connected, and hence, if we 
know the motion of one of them, we know the motion of the entire 
body. A revolving fly-wheel furnishes an illustration of plane 
motion. Each particle moves in a fixed plane perpendicular to 
the axis of the wheel. We may also consider the wheel to be made 
of thin plates joined rigidly together, which rotate in planes 
perpendicular to the axis of the wheel' As a second illustration 
consider a car travelling along a straight track. Each particle 
moves in a fixed vertical plane. A particle in one of the car 
wheels rotates about a point on the axle which is itself moving 
forward, and the motion takes places in one of the fixed planes 
described. 

167. External and Internal Forces. — In certain cases, as that 
of a body acted on by gravity only, the forces acting tend to 
produce the same acceleration in each of the particles, i. e., each 
particle would have the same acceleration even if it were not 
rigidly connected with the others. In most instances, however, 
change of motion in a rigid body is produced by the application 



Dynamics of a Rigid Body 287 

of forces to only some of the particles. Snch forces are called 
external forces. To illustrate, consider the motion of a railroad 
coach on a straight track. The pull of the drawbar, which is 
applied at one point, moves all the particles of the car without 
changing their relative positions. We know that it takes a force 
to accelerate any one of the particles and this force must originate 
from the drawbar, and be transmitted to the particle. Thus 
there must be forces acting between the particles of the body. 
These are called internal forces. They are the forces which resist 
the deformation of the body and keep it rigid. 

In dealing with internal forces we shall assume that the forces 
acting between two particles of the body have the same line of 
action and are equal and opposite. Thus in any summation of 
all forces acting on all particles of the body, all internal forces 
will cancel and leave only the external forces. In Chapter IV 
we have the truss, an example of a rigid body at rest, acted on 
by a system of external and internal forces. The loads and sup- 
ports constitute a system of external forces applied to certain of 
the particles; the stresses, induced by the loading in the various 
members of the truss, constitute a system of internal forces; 
further, the force exerted by any member on a pin is exactly 
equal and opposite to the force that the pin exerts on this member 
and has the same line of action. 

168. Translation. — If a body moves so that every line in it 
remains parallel to itself the motion is called translation. The 
particles will move in parallel curves. A common illustration 
is that of a block sliding down a plane with all its particles moving 
in straight lines, but in any motion of a compass box on a table 
the needle remains parallel to itself and its motion is translation. 
The particles of the needle describe curved paths. In transla- 
tion all particles of a body have at any instant the same velocity 
and the same acceleration. 



288 Elementaky Mechanics 

169. Motion of the Center of Gravity in Translation. — Sup- 
pose a body of mass M to have the motion of translation. Let dm 
be the mass of an elemental particle of the body, a its acceleration, 
and dF the resultant of all the forces acting on dm. Then 
dF = adm. As stated, it follows from the definition of transla- 
tion that all particles of the body have the same acceleration at 
any instant. Hence the several forces dF are parallel forces, 
acting parallel to the direction of the acceleration a, and the 
resultant of the system of external forces acting on the body acts 
in the same direction, and is 

^dF—\ adm = a j dm — aM, 
since the internal forces cancel in the summation. To find the 
point of application {x^, y^, z^) of the resultant we assume a set 
of coordinate axes (as in Art. 56) and by taking moments about 
them obtain 

XQ-a-M—^x-a- dm, y^-a-M^^y -a. dm, z^-a-M^^z-a- dm. 
Since a is constant for the body at any instant we may cancel it 
out. We then recognize these equations as the equations (4) which 
determine the coordinates of the center of gravity. 

Therefore, if a body has the motion of pure translation, the 
system of external forces acting on it is equivalent to a single 
resultant force acting at the center of gravity in the direction of 
the acceleration. Also, the center of gravity moves as if the 
whole mass were concentrated there and acted on by the resultant. 

170. Thus it follows that the solution of problems involving 
the translation of a rigid body is reduced to the consideration of 
a particle at the center of gravity of the body, whose mass equals 
that of the body, and which is acted upon by a force that is the 
resultant of all the forces acting on the body. The following 
problems can be solved by principles already studied in earlier 
chapters. In particular we can use the principles of work and 
energy as applied in Chapter IX to the particle. 



Dynamics of a Eigid Body 289 

Example 1. — What is the effective H. P. of a locomotive which 
gives a 500-ton train a speed of 15 m/h up a grade of 1 in 320, 
f rictional resistance being 12 pounds per ton ? 
Solution. — 

i; = 15x22/15=:22f/s. 
500 X 2240 X -jYo X 22 = work in ft.-lbs. in 1 sec. against gravity. 
12 X 500 X 22 = work in ft.-lbs. in 1 sec. against friction. 
.*. H. P.= (22 X 12 X 500 + 500 x 2240 X gfo X 22)/550 = 380. 

Example 2. — A 600-ton train moves down a grade of 1 in 50 
with a uniform velocity after power has been shut off, and upon 
reaching the level it runs half a mile before coming to rest. If 
the resistance remains constant find the velocity of the train on 
the grade. 

Solution. — Let v = velocity down the grade in miles per hour, 

i^=the resistance. 

Then K. E. in ft.-lbs. at the bottom of the grade is 

1/2 X 600 X 2240/32 X (22v/15) 2. 

Since there is just enough resistance to neutralize the accelerating- 
force of gravity, 

F= 600x2240x1/50. 

Substituting in the work-energy relation, 

we have 

600 X 2240 X 1/50 x 2640 = 1/2 x 600 x 2240/32 X (22 V15) ' ; 
.•.i; = 39.6m/h. 



Problems 

599. A freight car weighing 40 tons and moving 12 m/h is 
brought to rest on a level track in a space of | of a mile. Find the 
mean frictional resistance to the motion in pounds. 

600. A coasting party of five people, whose average weight is 
150 pounds, reaches the foot of a hill at a speed of 30 m/h. The 
sled weighs 50 pounds and //,= 1/10; how far will they travel 
along the the level ? 

19 



290 Elementaey Mechanics 

601. An engine is capable of producing 150,000 ft.-lbs. of 
energy per second ; find the greatest uniform velocity at which it 
can draw a train weighing 100 tons along a level track, the fric- 
tional resistance being 1/90 of the weight. 

602. A 500-ton train is moving up a grade of 1 in 95 against 
frictional resistance of 15 pounds per ton. What H. P. must the 
engine develop at the instant when the speed is 15 m/h if the 
acceleration of the train at that instant is ^ f /s^ ? 

603. A freight train, weighing 200 tons and travelling 20 m/h, 
runs into an engine weighing 50 tons and standing on the track. 
Find the velocity with which the broken engine will be forced 
along the track if e = ^. 

604. An elastic cord, which stretches 1 inch under a pull of 
5 pounds, is being used to raise a weight of 80 pounds through a 
height of 6 feet. If the work used in stretching the string is lost, 
what is the efficiency of this lifting device? 

605. A 25-ton freight car, moving with a velocity of 4 m/h, 
strikes a bumping-post. Assuming that the post absorbs none of 
the shock, find the amount of compression of the 50,000-pound 
spring of the draft rigging. 

606. A steam engine has a cylinder 2 feet long with diameter of 
10 inches. Steam with boiler pressure of 150 pounds per square 
inch is admitted to the cylinder during the first half -stroke and 
the valve is then closed. If the steam pressure varies according to 
the law P'V = const, find the work done during the forward 
stroke. If the flywheel is making 200 E. P. M., what is the H. P. 
of the engine ? 

171. Rotation. — When a body turns about an axis fixed in the 
body and fixed in space we have simple rotation. Each particle 
moves along the arc of a circle whose plane is perpendicular to the 
axis of rotation and all particles have at any instant the same 
angular velocity and the same angular acceleration. 

172. Equation of Angular Motion. — Let P be an element of 
mass dm of a rigid body which has the motion of simple rotation. 
Let OZ, Fig. 228, be the axis of rotation, r be the distance of 



Dynamics of a Rigid Body 



291 




Fig. 228. 



dm from the axis of rotation, /? the angular acceleration of the 
body, and dF the sum of the tan- 
gential components of the forces act- "Z^ 
ing on dm. The linear acceleration 
of P is then r(3, and the equation of 
motion of P is 

dF^rfBdm. 
Taking moments about OZ , we have 

rdF^r-^dm, 

and, integrating to cover the entire 
mass, 

where I is the moment of inertia 
of the body about the axis of rota- 
tion, and \rdF is the sum of the moments of the external 
forces acting on the body about the axis of rotation — it being 
remembered that the moments of the internal forces cancel in the 
summation. Denoting this moment by Mz, we have 

That is : The total moment of the external forces about any fixed 
axis of rotation is equal to the moment of inertia about that axis 
multiplied by the angular acceleration of the body. 

If we write the equation for the rotation of a body about a fixed 
axis, as I = Mz/^, we see that I is the measure of the moment of 
the force necessary to produce a unit angular acceleration in over- 
coming the inertia of the body. Hence the origin of the name 
" moment of inertia " given to I. It is analogous to the mass, 
M, in the equation for rectilinear motion, F /a — M, where M is 
the measure of the force necessary to produce a unit linear acceler- 
ation in overcoming the inertia of the body. 



292 



Elementary Mechanics 



173. 



Illustrative Examples 



Example 1. — A cylinder, weight 200 pounds and radius 15 
inches, is free to turn ahout its geometric axis. A fine cord is 
wound around it and a pull of 5 pounds is applied to the cord. 
Find the angular acceleration of the cylinder. 

Solution. — il/z = 5x5/4: lb.-ft. = sum of the moments of tlie 
external forces. 

7 = 200/32. A;- = 200/32.1. (5/4)- Ib.-iV 
Substituting in the above formula, 

5.5/4=1200/32.1.5/4.5/4^, 
/. ^ = 32/25 rad/sec." 
Example 2. — A solid cylindrical drum, weighing 175 pounds 
and 3 feet in diameter, is making 180 E. P. M. What tangential 
braking force will bring it to rest in 16 seconds ? 

Sohdion. — i^ . 3/2 = moment of external forces about axis of 
rotation. 

7 = 175/32.1. (3/2)^ 
^.10 = 0)=: 180/60 • 27r rad/sec = angular velocity. 
.'./? = 37r/8 rad/sec. 

1(1)2-^-, .-.F^: 4.8 lbs. 





Fio. 2?,0. 



Example 3. — A metal drnm, I feet in diameter and weighing 
288 pounds, has wound on it a line cord, to the end of which a 
weight of 30 pounds is attached. The drum revolves freely about 



Dynamics of a Rigid Body 293 

its axis, which is horizontal ; the radius of gyration about this axis 
is A; = 1.6 feet. Find the tension in the cord. 

Solution. — The equation of motion of the weight is (Fig. 229) 

30-^ = 30/32. a, .*. T = 30(l-a/32); 
also a = 2^. 

2T = moment of external forces, 
7 = 288/32. (1.6)^ 

Substituting in Mz=I/3 (Fig. 230), 

2^ = 288/32. (1.6)2^ = 288/32. (1.6)-. a/2. 
Solving simultaneously with the first equation, 

T = 25.81bs. 

Prohlems 

607. A grindstone, 3 feet in diameter and weighing 200 pounds, 
is making 45 E. P. M. when a tangential braking force of 3 pounds 
is applied at the rim. How long will it take for the stone to come 
to rest? 

608. A solid disk, weighing 980 pounds and 4 feet in diameter, 
is making 80 R. P. M. What tangential braking force applied at 
the rim will bring it to rest in 2 minutes 34 seconds ? 

609. A cylinder (with axis horizontal), 1 foot in diameter and 
weighing 100 pounds, has a fine cord wound on it, to the free end 
of which is attached a weight of 20 pounds. The cylinder is free 
to revolve about its axis. Find the linear and angular accelera- 
tions and the tension of the cord. 

610. A solid homogeneous sphere of radius 1 foot and weight 
2000 pounds is mounted on a horizontal weightless axle of 
diameter 4 inches, around which a small cord is wound carrying a 
weight of 50 pounds. Neglecting friction, find the angular 
acceleration of the sphere, the linear acceleration of the weight, 
the tension of the cord, and the velocity of the weight at the end 
of 5 seconds from rest. 

611. A grindstone, 4 feet in diameter and weighing 225 pounds, 
is making 150 E. P. M. A tangential friction of 6 pounds is 
applied at the rim. How many complete revolutions will the 
stone make before coming to rest ? 



294. 



Elementary Mechanics 



174. The Compound Pendulum. — A body which rotates about a 
fixed horizontal axis under the action of its weight is called a com- 
pound pendulum. 

Pass a plane through G, the center of gravity of the body, 
perpendicular to the axis of rotation, and let the section be 

represented by Fig. 231. The axis of 
rotation is a line through perpendicu- 
lar to this plane. Let IF = the weight 
of the body, Jvq = the radius of gyration 
about 0, iv = the distance OG, ^ = the 
angle OG makes with the vertical. Since 
we may replace the gravitational forces 
by their resultant, W, acting through G, 
the moment of the external forces about 
the axis of rotation is WL sin 0. Then 
by the equation of angular motion, 

g '"^^ ' dV-' 




Fig. 231. 



^YL^ille=l(i 



whence 



dP 



— gL sin 

h' ■ ' 

Placing the constant L/Jvq- = l/l, we get - 

d-0 _ —g sin 

dr- ~ I ' 
which is the differential equation that was obtained for the motion 
of the simple pendulum in Art. 159. Hence a compound pendu- 
lum will oscillate about an axis in exactly the same manner as a 
simple pendulum whose length is equal to the squared radius of 
gyration of the body about the axis of rotation divided by the 
distance of the center of gravity from the same axis. 

175. Centers of Suspension and Oscillation.— The point is 
called the center of suspension. The point 0' , obtained by 



Dtxamics of a Eigid Body 295 

measuring the length A-,,- L from along OG, is called the center 
of oscillation. 

Let A-G- = the radins of gyration of the body about an axis 
through the center of gravity parallel to the axis of rotation, then 

and 
therefore. 

Since L and 1 — L enter the expression in exactly the same way, it 
follows that if 0' were taken as the point of suspension, would 
be the center of oscillation. Therefore, in a compound pendulum 
the centers of suspension and oscillation are interchangeable 
without altering the period. 

Example. — Find the time of Tibration for a small oscillation of 
a plank, -i feet by 1 foot by 1 inch, about an axis perpendicular to 
the 4 foot by 1 foot face through the middle point of the 1 foot 
edge. 

.Solution. — 
lY = l-o- + '2- = 6b/l'2 it.- 
I=6b 24: ft, = length of equivalent simple pendttlitm. 
f = 7r\Ty=7rA 65/24- 1/32 = 0.91 sees. 
The center of oscillation is 17/24 ft. from the center of gravity; 
therefore, if the axis were moved to a point 17/21: ft. from the 
center of gravity, the time of vibration would not be changed. 

Frohlcms 

612. A solid right circular cylinder, radius (7, oscillates about a 
horizontal axis which is parallel to the geometric axis of the 
cylinder and at a distance of Sa, 2 from the latter. Find the time 
of a small vibration. 

613. A thin imiform rod, 1 feet long and 1 inch in diameter, 
oscillates about a horizontal axis, 1 foot from one end and parallel 
to a diameter of one base. Find the length of the equivalent 
simple pendulum. 



296 



Elementary Mechanics 



614. A cylindrical bar of 4 inches radius and 2 feet long is 
suspended from an axis which coincides with a diameter of one 
of its ends. If it is slightly disturbed how many oscillations will 
it make in a minute ? 

615. What must be the ratio of the radius of the base to the 
height of a right circular cone, in order that the center of oscilla- 
tion may be in the base when the center of suspension is at the 
vertex ? 

616. A cast-iron sphere whose radius is 6 inches vibrates as a 
pendulum about a tangential line as an axis. Find the period of 
vibration and the length of a simple pendulum having the same 
period. Locate the center of oscillation. 

617. A board 4 feet by 1 foot by 1 inch vibrates about an axis 
perpendicular to the 4 foot by 1 foot face through a point 
18 inches from the center. Find the time of vibration for small 
oscillations and the length of the equivalent simple pendulum. 

176. The Torsion Balance. — A torsion balance or pendulum 
consists of a rigid body suspended by a thin rod, the axis of which 
passes through the center of gravity of the body. The rod is 
rigidly joined to the body at the point of support and held 
rigidly at the other end also. 

////////////////////////////////// . Suppose we have a torsion pendulum 
consisting of a suspended disk as shown 
in Fig. 232. Let OU be a radius marked 
on the upper surface of the disk at the 
position of rest. If the disk is turned 
through an angle 6q (OR will then take the 
position OR') the rod is twisted, and when 
the disk is released the rod exerts a twist- 
ing moment on the disk tending to return 
it to the initial position of rest. By the 
equation of angular motion, ]\lz — l(i. But 
in Art. 109 it was shown that the twisting 
moment exerted by the rod is proportional to the angle of 
twist. Hence we may write 




Fig. 232. 



Dynamics of a Rigid Body 297 

which is recognized as the equation of simple harmonic motion 
when we substitute 6 for s and k/I for Jc- in the equation of 
Art. 133, and proceeding as before we find 

arc sin ( 6 /Bo ) = 7r/2 - Vk/I t, 

and hence the time of one swing from the extreme left to the 
extreme right is 

If we know the twisting moment, M-^, exerted by the rod when 
twisted through any given angle 6-^, we have M^ — kO^, and hence 

This equation gives the time of vibration independently of the 
initial displacement. 

Example 1. — A uniform metal disk, 9 inches in diameter and 
weighing 6 pounds, is suspended from its center by a wire so that 
its plane is horizontal, and then twisted. What is the time of a 
complete oscillation if a twisting moment of 2 lb. -ft. causes a 
deflection of 12°? 

Solution. — 

Mi = 21b.-ft. 
^i = 12x7r/180 = 7r/15rad. 
7 = 6/3 2.2 xix (f)Mb.-ft.^ 
i = 2-77 V7^i/ilfi = time of a complete vibration. 
.-. J^ = 27rV^^XiXf X|Xy5Xi = 0.233 sees. 

Example 2. — The moment of inertia of a torsion balance is 
6300, where units are pounds and inches, and its time of vibration 
is 20 seconds. A disk L is fitted to the balance so that its center 
of gravity lies in the axis of suspension. The time of vibration is 
then found to be 30 seconds. Find the moment of inertia of L. 

Solution. — Let I = the moment of inertia of the disk. Substi- 
tuting in the formula. 



t = 7rVl/k; 



298 Elementaey Mechanics 

before the disk is attached 

20=:7)-Vl)o00/A-; 
after the disk is attached 



30 = 7rV(6300 + /)A-. 
Dividing one by the other, 



| = V()300/(G300 + /). 
Solving, 

7 = 7875 Ib.-iu.-. 

rrohlenis 

618. A fly-wheel of 3 tons is fastened to one end of a shaft, the 
other end of which is fixed, and the torsional rigidity of which is 
such that it twists 0.4° per ton-foot of twisting moment applied 
to the flywhoel. If the radius of gyration of the flywheel and 
shaft combined is 3 feet, find the number of torsional vibrations 
per minute the wlieol would make if slightly twisted and then 
released. 

619. The moment of inertia of a torsion balance fitted with a 
platform is 7500 in Ib.-ft units, and the time of vibration is 
25 seconds. A body L placed on the platform with its center of 
gravity in the line of suspension, and the time of vibration is 
observed to be 30 seconds. Find the moment of inertia of L. 

177. Kinetic Energy of Rotation and the Work-Energy Rela- 
tion. — In a rigid body, let dm be the mass of an element whicb 
lias linear velocity v and is at a distance r from the fixed axis of 
rotation. The kinetic; energy of this (dement is } • dm ' v", and 
that of the entire body is the sum h\v'~-dni. Since v = no, 
where w is the angular velocity of the body, we have 
^i \ v~ ■ dm = I \ /•"(.)■- • dm = i • / • or, 

that is, Tlie hinclic eiicnjij of a bodij rolaling nhoul a fixed 
axis is equal to iJie 'moment of inertia of the body about tliis a.vis 
viultiplied by one-lialf the square of the angular velocity of the 
body. 



Dynamics of a Rigid Body 



299 



Let dF be the sum of the tangential components of all forces 
acting on the element of mass dm. By the equation of angular 
motion 

Sr . dF = Mz = I ' ^ = 1 • d'e/dt\ 



v/hich we may write 
{\r-dF) ' 



iI{2'd'-0/dt^-de/dt.dt). 



But io = dO/dt and d<^ = d~0/dt^ • dt. Substituting these values, 
we have 

{\rdF)de = iI'2oj'd<o; 



integrating, 



^{^rdF)dd = iIio' + C. 



work done by dF as the body 

ds. 



From Fig. 233 it is seen that the 
rotates through an angle dd is dF 
but ds = r- d6, and the total work done 
by all the forces acting on the body is 
found by the summation \ {^rdF) dd. 
If coq denote the initial angular veloc- 
ity, then when w = a>o, \{\rdF) dO — O 
(no work having been done) and 
therefore (7= -iK'- Thus 

S{SrdF)de = lI{o.^--<o,^), 

which equation states that : The ivorh 
done hy the external forces (since the 

internal forces cancel in the summation) on a body rotating about 
a fixed axis is equal to the change in the Icinetic energy of the 
body. From this equation it also appears that if the body has an 
initial angular velocity w, the work it can do before coming to 
rest is ^Ico^, and hence J/a>- is the kinetic energy of the body rotat- 
ing about a fixed axis with angular velocity w. 




Fig. 233. 



300 



Elementary Mechanics 



Example 1. — A cylinder, of weight 200 pounds and radius 15 
inches, rotates about its geometric axis, making 120 E. P. M. A 
tangential force of 5 pounds is applied to its lateral surface. How 
many turns will the cylinder make before coming to rest ? 

Solution. — The work done against the tangential force in one 
revolution is 27r • 5/4 • 5 ft.-lbs. For n revolutions the work done 
is n. 25/2. TT ft.-lbs. 

120/60- 27r = the angular velocity in rads. per sec. 

Setting the total work done = the change in K. E., 

n . 25/2 . 7r = i . 200/32 . i • (5/1)2 . (47r)2; 
.'. 71 = 9f I revolutions. 

Proljlems 

620. A solid wheel, 2.6 feet in diameter and weighing 140 
pounds, is making 50 E. P. M. Neglecting friction, find the 
amount of work in ft.-lbs. which must be done to increase its 
angular velocity to 70 E. P. M. 



!00 lbs 




Fig. 234. 



621. A solid cast-iron drum, of radius 3/2 feet and thickness 
1 foot, is making 200 E. P. M. when a brake is applied and it is 
brought to rest. The brake is applied from below by a 100-pound 
force acting at the end of an 8-foot hinged beam (Eig. 234) which 
the drum touches 4 feet from the hinge. The coefficient of friction 
between the drum and the brake is J. The radius of the axle is 
1 inch and the coefficient of axle friction is 1/20. How many 
revolutions will the drum make before coming to rest ? 

622. A fly-wheel, weighing 15 tons and of diameter 20 feet, is 
making 60 E. P. M. Find the energy stored up. If the axle of 



Dynamics of a Eigid Body 301 

the wheel is 1-i inches in diameter and the coefficient of axle fric- 
tion is f, when the wheel is disconnected how many revolutions 
wdll it make before coming to rest ? 

623. Find the kinetic energy of rotation of the earth, consider- 
ing it a uniform sphere of density 5.6 and of diameter 8000 miles. 

624. To control an engine against its own variations it is neces- 
sary to call upon the fly-wheel for 60,000 ft-lbs. of energy, and at 
the same time a change of velocity from 160 E. P. M. to 140 
E. P. M. is allowable. If the wheel is 10 feet in diameter, com- 
pute its weight in tons. 

178. Power Transmitted by Shafts. — In a shaft we have a 
cylinder rotating about its geometric axis and transmitting 
energy. Let t be the length of the crank arm and F the mean 
force acting on it through one revolution. Then the work done in 
one revolution is 2-^1) - F — 2tt - T, where T is the mean twisting 
moment. (See Art. 109.) To find the H. P. transmitted, we 
must express the work done in ft.-lbs. per minute and divide by 
33,000. 

Example 1. — A solid steel shaft, diameter 6 inches, is used to 
transmit power. It is to make 160 E. P. M. and the allowable 
stress is 5000 pounds per square inch. What H. P. may be 
transmitted ? 

Solution. — 

T = g7rrV2 = 5000 . 22/7 . 372 = 212,000 Ib.-in. 

H. P. = 27rrV33000 = 2 . 22/7 • 70714/12 • 160/33000 = 539. 

Example 2. — The resistance of a twin-screw vessel at 18 knots 
is 44,000 pounds. At 95 E. P. M. what will be the twisting 
moment on each shaft ? What is the H. P. ? 

Solution. — Let r = the mean twisting moment on one shaft in 
ton-ft. Then 

2 . 27rT . 95 = the work done on the shafts in one minute in ft.-tons. 
18/60. 44000/2240. 6080 = work done against resistance in ft.- 
tons in one minute. 

H. P. 33000/2240 = work done in one minute in ft.-tons. 
Equating these quantities, 

2 . 27rT . 95 = 18/60 • 44000/2240 • 6080 = H. P. 33000/2240 ; 
.'. r = 30 ton-ft., H. P. = 2432. 



302 



Elementaey Mechanics 



Problems 

625. What H. P. can a steel shaft 6 inches in diameter transmit 
at 100 E. P.M.? 

626. Find the diameter of a solid shaft to transmit 9000 H. P. 
at 140 E. P. M., stress allowed being 10,000 pounds per square 
inch, and maximum twisting moment 3/2 the mean. 

627. Find the size of a hollow steel shaft to replace the shaft 
of the preceding problem, the internal diameter being i% of the 
external. What is the saving in weight for 60 feet of shafting ? 

628. The pitch of a screw propeller is 14 feet, and the twisting 
moment on the shaft is 120 ton-in.; if the mean diameter of the 
thrust-bearing rings is 15 inches and the coefficient of friction is 
.05, find the thrust and the efficiency of the thrust bearing. 

179. Translation and Rotation : Plane Motion. — In the general 
case of plane motion the center of gravity of the body moves 

(translates) in a plane and the 
particles of the body rotate 
about an axis through it per- 
pendicular to the plane. Let 
the coordinates of G^ be (^'^ ^), 
the coordinates of any particle 
P, of mass dm, be {x, y) and 
{r, 6) with respect to a set of 
axes through G parallel to the 
first set. (Fig. 235.) Then 

a; = :X- + f cos 6, 
y — y-\-r sin 6, 




Fig. 235. 



Differentiating with respect to t, 

Vx = dx/dt — r sin ^ dO/dt, 
Vy — dy/dt + r cos ^ dd/dt. 
Differentiating again, 

ax = d^x/dt" — r sin d d'^d/dt- — rcos6- 
ay^d^y/dt"" -f r cos (9 d-B/df - r sin 6 • 



(de/dt)\ 
{de/dty. 



(6) 



Dyxamics of a Eigid Body 303 

The resultant of the forces acting on dm is determined by the 
components 

dX = a.rd m . (71' = a-ydm ; 

and the resultant of all the external forces acting on the body — 
since the internal forces cancel in the summation — by 

A' = \ ajrdni, I' = j aydm . 

Substituting from equation (6), 

X = ( d-l\ dt- . dm - \ d-O/di- • r . sin ^ • dm - 

\{dO'dt)-' r-cose- dm: 
Y=\d-T//di- . dm + \d-e^ di- • r • cos • dm - 

\{dO dt)-.r.sme-dm. 

But ( r . sin ^ . dm = \ ij - dm = 0, ( r cos 6 . dm = \x - dm = 0. since 
the origin is at G. (See Art. 56.) Therefore, since all the deriva- 
tives are constant for the integration. 

X = j d-J-/dt- .dm=2I . d'!; dt-, 

T=\d-y/dt- . dm =M . d^y/dt^, ^ ' 

showing that the center of gravity of a body in plane motion 
moves as if the entire mass of the body were concentrated at that 
point and all the external forces were applied to it parallel to 
their original directions. 

Taking moments about G of the forces acting on dm, we have 

dJIz = — r . sin ^ . a^- • dm -f ;• . cos -ay dm, 

and, integTating, we find the sum of the moments of all the forces 
actino: on the bodv: 



J/z = \ - /• • sin ^ . d-I/df- . dm + ( sin- 6 • d-d, di- . /•- . (7 



/;? 



-r \ sin e • cos e . ((7^^ (77) - r . dm + j r • cos • d-J//dt- . (7/;? 
+ Jcos- e . t7-^/(77= . r • dm - jsin (9 cos 6 {dO/dt) - • F . dm. 



304 Elementaey Mechanics 

The first and fourth integrals are zero for the reason given above, 
the third and sixth cancel, and the equation reduces to 

Mz = d~e/dr~^r^ . dm = Ig/3. (8) 

where Iq is the moment of inertia of the body about the axis of 
rotation through G. 

From equations (7) and (8) we see that when a system of 
coplanar forces acts on a rigid body (in Art. 46 such a system was 
shown to be equivalent always to a single force and a single 
couple), the motion of G is governed by the magnitude and direc- 
tion of the resultant force and is independent of its point of appli- 
cation, while the rotation of the body about the axis through the 
center of gravity is governed by the moment of the resultant 
couple and is independent of the motion of the center of gravity 
and the magnitude and direction of the resultant force. 

The kinetic energy of a body in plane motion is the sum of the 
kinetic energy of translation and the kinetic energy of rotation. 
Thus if V denote the linear velocity of G we have 

The work done by the body is found by taking the sum of the 
following equations (see Art. 164) : 

Xdx = M/2' (va^'-v^,'); 

Ydy = M/2-{v/-Vy,') ; 

Mzde = I/2- ((o--a>o-). 

Thus, WoTk = M/2.(v^^-Vo^) +i-^G- {o^^-c^o"), or 

The work done by a body in plane motion is equal to the change in 
the kinetic energy of translation plus the change in the kinetic 
energy of rotation about the axis through the center of gravity. 




Dynamics of a Eigid Body 305 



180. Examples 

Example 1. — A right circular cylinder, weight 10 pounds and 
radius 3 inches, rests on end on a smooth table. A fine cord is 
wound around the cylinder and a force of 1 pound applied to the 
end of the cord. Determine the mo- 
tion of the cylinder. 

Solution. — Applying the two equal 
and opposite forces F' and F" (Fig. 
236) at G, we see that F' is the re 
sultant force which will cause the 
cylinder to translate. Thus, 

a = 32/10 = 3.2 f/s-. 

The couple F' F" has a lever arm of 
3 inches, and we find the angular ac- 
celeration from the equation M — Iq^. 

/?= (1 X 3/12)//g = 3/12 . 64/10 . 12/3 . 12/3 = 25.6 rad/sec.^ 

The center of gravity of the cylinder will move with an accelera- 
tion of 3.2 f/s^ and the cylinder will rotate about its geometric 
axis with an acceleration of 25.6 rad/sec^. 

Example 2. — A cylinder, 2 feet in diameter and weighing 20 
pounds, rolls down a rough plane 48 feet long and inclined 30° to 
the horizontal. If it starts from rest what is the angular velocity 
at the bottom ? 

Solution. — 

20 . 48 . J = work in ft.-lbs. done by the cylinder; 

(Si = v/r — v/\ — angular velocity in rad. per sec. ; 
\ . 20/32 . i;^ = gain in kinetic energy of translation; 
^•20/32 • -1(0- = gain in kinetic energy of rotation. 

Therefore 

20 . 48 . J = i . 20/32 . i;2 + 1 . 20/32 . f 72 . a>2 
= |. 20/32. i;^ 
.*. i;=32 f/s, (o = ?;/r=32 rad/sec. 

20 



306 Elementaey Mechanics 



ProiUms 

629. A cylinder rolls down a 30° incline. Find its velocity and 
the distance it has rolled at the end of 9 seconds. What is its 
velocity after it has rolled 18 feet from rest? 

630. A solid 30-pound sphere rolls down a 30° incline in 
7 seconds. Find the length of the incline. 

631. A solid sphere, a disk, and a hoop start rolling from rest 
at the same time at the top of an incline. It what order will 
they reach the bottom? 

632. A hoop rolls in a vertical plane. Show that the energy of 
rotation is one-half of the total kinetic energy. 

633. The kinetic energy acquired by a sphere in sliding (with- 
out rolling) from rest down a smooth plane, is to that acquired 
by an equal sphere rolling (without sliding) down a rough plane 
by the same inclination and length, as 7 to 5. 

Review Pi^ohlems 

634. Two solid cylinders, weights 400 pounds and 800 pounds, 
radii 1 foot and 2 feet respectively, are mounted on an axle 
coinciding with their geometric axes. Weightless cables are 
wound around these cylinders — the cable on the smaller being 
attached to a weight of 250 pounds, that on the larger to a weight 
of 300 pounds. If the system is initially at rest and friction is 
neglected, find: (a) the velocity of the 300-pound weight when 
it has moved 10 feet; (b) the angular velocity of the cylinders at 
that time; (c) the angular acceleration Of the cylinders and the 
linear accelerations of the weights; (d) the time of descent. 

635. A car weighs 40,000 pounds, the eight wheels weigh 4000 
pounds, and the speed is 30 m/h. Find the kinetic energy 
stored up. 

636. Show that " there is more energy stored in a ton of car- 
wheel than in a ton of car-body.'' (K. E. Gazette.) How much 
when the speed of the train is 40 m/h ? 

637. A bullet weighing 1 ounce is fired horizontally into a box 
of sand (inelastic) weighing 20 pounds, and remains imbedded in 
the sand. The box is suspended by a string attached to a fixed 
point 4 feet from the. center of the box of sand. The impact of 
the bullet causes the box to swing aside through an angle of 42°. 



Dynamics of a Eigid Body 



307 



w 

60 lbs 



E 



Fig. 237. 



Find the velocity of impact of the bullet, the kinetic energy lost 
in impact, the greatest tension in the string. 

638. A weight W rests ©n a smooth table and is connected with 
a second weight i? by a cord passing over a smooth pulley A. 

(Fig. 237.) When the motion 
starts, H is 7/2 feet above the 
floor. Find the velocity when H 
strikes the floor. 

639. A solid cylinder, 18 
inches in diameter and weighing 
300 pounds, which revolves 
freely on its axis without fric- 
tion, has a fine cord wound on it 
and a weight of 20 pounds attached to the end of the cord. The 
support under the 20-pound weight is removed and the weight 
descends under the action of gravity. What is the kinetic energy 
stored in the cylinder when the cord has caused the cylinder to 
make 3 complete revolutions ? 

640. A twin-screw steamship of 22,000 H. P. runs 3300 miles 
in 6 days. Find the resistance to the motion. At 95 E. P. M. 
what will be the mean twisting moment on each shaft ? 

641. A weight of 20 pounds rests on a rough table (fx = ^). A 
is a smooth pulley over which 
passes a string connecting W with 
the weight P = 10 pounds. The 
string is horizontal from W to A. 
Find the tension of the string. 
(Fig. 238.) 

642. A 200-pound weight is 
projected down AB with an initial 
velocity of 24 f/s at A. AB is 



w 

20 



k) 



EM 



Fig. 238. 



A 










^^ 


60 


\<^ 






j^^"^^ 


100' 


H 




nB^ 


^'^ 








eo' 




Fig. 


239. 





N 



308 Elementary Mechanics 

rough (/A = 1/10) ; no shock occurs at B, and the weight continues 
up the rough grade BC (/x = l/20). Find the K.,E. of the weight 
when it is half way up BC and still rising. (Fig. 239.) 

643. A cylinder of weight W rests on a rough horizontal table 
with axis parallel to and s feet from the edge. A cord is wound 
around it, passes over a smooth pulley at the edge of the table and 
is attached to a weight H. Find how long it will take the center 
of the cylinder to reach the edge of the table. 

644. Find the work done by a locomotive that changes the 
velocity of a train weighing 200 tons from 25 m/h to 50 m/h in 
2 minutes, the frictional resistance being 10 pounds per ton. 

645. A wheel is making 200 K. P. M., and after 10 seconds its 
speed has fallen to 150 K. P. M. If the angular retardation be 
constant, how many more revolutions will it make before coming 
to rest ? 

646. If it takes 600 useful H. P. to draw a train of 335 tons 
up a grade of 1 in 264 at a uniform speed of 40 m/h, estimate the 
resistance per ton other than that due to ascending against gravity, 
and find the uniform speed on the level when developing the above 
power. 

647. A uniform circular plate, 1 foot in diameter and weighing 
4 pounds, is hung in a horizontal plane by three fine parallel cords 
from the ceiling, and when set in small torsional vibrations about 
a vertical axis is found to have a period of 3 seconds. A body is 
laid diametrically across it and the period is found to be 5 seconds, 
the weight being 6 pounds. Find the moment of inertia of the 
body about the axis of oscillation. 

648. The mass of a flywheel may be assumed to be concentrated 
in the rim. If the diameter is 7 feet and the weight 5/2 tons, 
estimate its K. E. when running at 250 E. P. M. If the shaft is 
6 inches in diameter and the coefficient of friction of the shaft 
in the bearings is .09, find the number of revolutions the flywheel 
will make before coming to rest. 

649. If a hammer whose weight is \Y , falling through a heiglit 
li, strikes a pile of weight P, and drives it into the clay or sand, 
show that the energy available for penetration is W / {W -\-P)' ^Vh, 
supposing no rebound of the hammer. If W = -J ton, P = -J ton, 
h = 10 feet, and the pile is driven 6 inches into the ground, find the 
mean resistance of the ground. 



Dynamics of a Eigid Body 309 

650. A projectile has a kinetic energy of 1,670,000 ft.-lbs. at a 
velocity of 3000 feet per second. Later its velocity is only 2000 
feet per second. How mnch K. E. has it lost? 

651. An 80-ton gun discharges an 800-ponnd shot with a 
velocity of 2200 f/s. If the recoil is resisted by a constant pres- 
sure of 15 tons, how far will the gun recoil ? 

652fc Calculate the kinetic energy of rotation of a projectile if 
its weight is 12 pounds, its radius of gyration is 0.75 inches, and 
its speed of rotation is 500 R. P. S'. 



310 Elementary Mechanics 



ANSWERS 

2. 7, 21° 47'. 
7. 7.25 mi., ¥. 88° 05' E. 

9. (a) 14.53, 80° 06', 58° 55', 147° 04'; (b) 9.G6, 37° 05', 
108° 05', 121° 09'. 

10. (a) 47.42, 335° 03'; (b) 16.76, 72° 39'; (c) 39.56, 163° 
51'; (d) 30.15, 264° 17'. 

11. V3 perpendicular to second displacement. 

12. Vl3. 

13. -S/VS". 

14. 46.35. 

15. 6.32, 161° 34'. 

16. 13 lbs., 22° 37'. 

17. V2lbs., S.W. 

18. P lbs. perpendicular to given force P, 

19. 20 lbs., 30° 31' with 13 lb. force. 

20. 2 lbs. 

21. 18.03 lbs., 17° 26'. 

22. 60°. 

23. 6.46 lbs. 

24. 150° 19'. 

25. 3 lbs., 4 lbs., 5 lbs. 

26. 11.5 lbs., 23.1 lbs. 

27. 3 lbs. 

28. 2.5 tons. 

29. 64.76 lbs. 

30. 5 lbs., 13 lbs. 

32. T^Wl/2V¥^\ • . 

33. P = 4/3 W, T=:5/3Tf. 

34. C=125 lbs., r = 75 lbs. 

35. (7 = 3.86 tons, T = 2.83 tons. 



Answees 311 



36. F=lPi3.ne, 

37. 267.5 lbs, 

38. 74.4°. 

39. 2.89 ft., 0.85 ft. 

40. 17.32 lbs. 



41. wy2Rh-hy{E-ii), 

42. 112 lbs. 

44. 62.6 lbs., 42° 11'. 

45. 14.15 lbs., 77.95°. 

46. 500 lbs., 300 lbs. 

47. (a) 9 lbs., (b) 89.55 lbs., (c) 1/10. 

48. 12.68 lbs., 61.96 lbs. 

49. 53° 08', 160 lbs. 

50. (a) TF = 25.98 lbs., iA' = 15 lbs.; (b) P = 6.65 lbs., J\^=r 
5.94 lbs. 

51. 20 tons. 

52. 19.62 lbs. 

53. tairV. 

54. 21.05 lbs. 

59. 6.52 lbs. 

60. 30°, 0.577. 

61. i, 10 VlO lbs. 

63. i^-^HL(^±^. 

cos a + /A sin a 

64. 5 VI lbs., 5 lbs. 

65. (1) P/V2; (2) 12/13 P. 

66. 50 lbs. 

67. 36.60 lbs., 44.83 lbs. 

72. (b) 1:1: V^. 

73. V^ : 1 : 2. 



312 Elementary Mechanics 

74. 90° between 5TF and 12TF, 112° 37" between 12W and 
13F, 157° 23' between 13T7 and 5Tf . 

75. 226° 34' and 151° or 133° 26' and 209°. 

77. 40 lbs. 

78. 104° 29'. 

79. 2V3; V3 or 6, 3V3. 

81. 34 lbs., 20 lbs. 

82. 525 lbs., 450 lbs. 

83. 3 lbs., 13 lbs. 

84. (a) P = R = 5.77 lbs.; (b) ^ = 10 lbs., i? = 8.66 lbs.; 
(c) Q = 3.92 lbs., i^ = 5.28 lbs. 

85. (a) i^ = 30 lbs., Q = 26 lbs.; (b) P = 2570 lbs., 7^ = 3760 
lbs.; (c) P = 9 = 1.45 lbs. 

86. P/2, P/2. 

87. P:Q=V^:1. 

88. 5V2 lbs. in leg of rt. triangle; 10 lbs. in cord inclined at 
30° ; 5 ( V3~- 1) lbs. in hypotenuse. 

91. 2930 lbs., 2070 lbs. 

92. 282 lbs. 

95. 10 lbs., 6 lbs. 

96. 5759 lbs. 

97. lOVlO lbs., tan-i 3. 

98. (a) 16 lbs. 10" from 10 lb. force; (b) 4 lbs. 27" from 
10 lb. force. 

99. 2 1bs., J;=-1.5", ^ = 2.5". 

100. IOV2 lbs. 

101. 31.4 lbs. 

102. A 29.6 lbs., B 49.6 lbs., C 20.8 lbs. 

103. 204.0 lbs., 186.2 lbs., 146.2 lbs. 

104. 18 Ib.-ft. 

105. 4 lbs., 2 lbs., 20f lbs., 4^ lbs. 

106. 4f ft., 6 ft., 224 ft., 40i ft. 

107. 17.2 lbs. 



Answees 313 



108. 


81 Ib.-ft., 9 ft. 


109. 


16 lbs., 17f ft. from 0. 


110. 


15 ft. 


111. 


4 ft. 


112. 


4f ft. 


113. 


4 lbs., 8 lbs., 12 lbs. . ■ 


114. 


IS^V lbs. 


115. 


20 lbs. 20.8 ft. from 0. 


116. 


27 in. 


117. 


2V'2Pa. 


118. 


130V3 lbs. 


119. 


36 lbs., 27 lbs. 


120. 


130 1bs., lOVlOSlbs. 


121. 


T = 21.7 lbs., P = 33.4 lbs., ^ = 19.7° with vertical. 


122. 


BC = S4:6 lbs. (T), Eeaction at .4 = 630 lbs., ^ = 47° 50' 


with horizontal. 


123. 


T=10 lbs., R = 22A lbs., ^ = 63° 26' with horizontal. 


124. 


1167 lbs. (C). 


125. 


3000 lbs. (C), 


126. 


1.20 tons (T), 


127. 


410 lbs. (C). 


128. 


Horiz. 9000 lbs. (C) , diag. 707 lbs. (T) . 


129. 


Upper 6184 lbs. (T), micl. 2500 lbs. (T). 


130. 


Aa = Ba=DOOlhs. (C), Cfl = 400 lbs. (T). 


131. 


^a = 2125 lbs. {C),ab = 0. 


132. 


56 = 1200 lbs. (C), C6 = 1500 lbs. (T). 


133. 


Aa=2600lbs. (C),a6 = 560 1bs. (T), Da = 24.00 Ihs, (T), 


134. 


Aa = 20tons (C), 56 = 16 V2 tons (C) , ha = CD = 7 tons 


(T), Da = Cb =:zl6 ton (T). 


135. 


al) = CD = 4.200 lbs. (T), Da = 3375 lbs. (T), 56 = 5625 



lbs. (C) 



314 Elementaey Mechanics 

136. 5^ = 3900 lbs. (T), CE = 6000 lbs. (C), Ah=Dh^'^'^QO 
lbs., i>F=3300 lbs. 

137. AC=17 tons (C), Eh=Ah=24: tons, Ey=10 tons, reac- 
tion at B on BD = 9A3 tons, inclination with BD 212°. 

138. AO = 8081 lbs. (T), J5'f = 2000 lbs., ^£r=10,000 lbs., 
FD = 14:,U2 lbs. (C), i?F=13,714 lbs., Bjj=5714 lbs. 

139. 5^=5000 lbs., Bv=8000 lbs., (7^=5000 lbs., Ck = 1500 
lbs., Dh= 5000 lbs., Dy = 2000 lbs. 

140. 200 lbs., 200 V3" lbs., 30°. 

141. 200 lbs., 250 lbs., tan-^ f. 

142. «& = 300V2lbs. (C). 

143. R^ = R^ = 1G tons, Z)a = Z7c = 50.6 tons (C), Aa = Bl) = 
Cc = 48 tons (T), Dh = ^S tons (0), ah = hc = 16 tons (C). 

145. 55 = 40.2 tons (C), A& = 32.8 tons (T), aJ) = 32.H tons 
((7), Aa=56.8 tons (T), Ca = 28.4 tons (C), 17 = 49,2 tons, 
5(7=61.2 tons. 

146. ila = 1058lbs. ((7), i7a = 908 lbs. (T), a5 = CZ) = 360 lbs. 
(T),5& = 1135 1bs. (C). 

147. &e = 25 tons (C), 6^ = 0, c^ = 30 tons (C). 

148. 6c = 0, t^e = 1270 lbs. (T), 5e=,635 lbs. (T). 

149. hd=4:i tons (C), Dc^ = 12 tons (T). 

150. 55 = 9| tons (0), &^ = 4i tons (7), t?c = 8 tons (C). 

151. 1^=1700 lbs. ((7),C6 = 1500 1bs. (T). 

152. Eeaction at C= 1250 lbs., reaction at B and D = 5150 lbs, 

153. Eeaction at (7 = 2063 lbs., at B and D = 5d70 lbs. 

155. h/3. 

156. 2a{siiiA)/3A. 

157. 3h/4 from vertex. 

158. 3a-/8 from base. 

159. ^ = 5a/6. 

160. 5; = a(sinA)/l. 

161. 2V3 from vertex. 



Answers 315 

162. 2h/3 from vertex. 

163. £- = Y/ = 4a/3. 

164. (18/5, 18/5). 

165. a/4 from base. 

166. (2a/ 0, 2a/ d). 

167. (4fl/5, 0). 

168. (5a/ 6, 0). 

169. Bisects line joining centers of bases of zones. 

170. 2(n + 2)a/(n + 4). 

(V 1 + cos^ X sin X dx \ 
% ^' r. I, (-A -/8), (V^, 0), 

^5 Vl + COS^T^T / 

(7r/2,.0). ' 

172. (7ra/4, a/2, 7ra/8). 

173. 3a/4. 

174. 50«/63. 

175. (6 A, 6/,). 

176. 15/194 in. from center. 

177. Center of hole is 16" from center of disk. 

178. 3080 miles. 

179. li(2-V2). 
.Qf. h 2h + a 

^^^' 3"^+y 

181. 1.57 calibers from base. 

182. 5.12" from bas'e. 

183. 5.54" from base. 

184. ia. 

186. S = 7ra'^/1, 7 = 7raV4. 

187. S=4:7r^ah,V=27r'a'b, 

188. 2a/7r, 4a/37r. 

189. 76.8 TT. 

190. 47ra&V3. 



316 Elementaky Mechanics 

191. 77V/4. 

192. 7/3 by integration. 

193. 118.8. 

194. 6.802 TT. 

195. 20.4. 

196. 3523.3 cu. ft. 

197. 15.24 sq. in., 2.96 in. from base. 

198. 14,180. 

199. 49,405 cu. ft. 

200. 450 sq. ft., 14,320 en. ft., 17.73 ft. 

201. h]i'/12 = M]i^/6. 

202. 3/10 Mr\ 

203. iMa^. 

204. liTraMfTra^ 

205. 21/512 TT^^ 

206. i¥6-/4. 

207. 4/15 aJ)\ 4a^h/7. 

208. 7ra&V6. 

209. 9369 in.% 94.64 in.^ 

210. 1120/3 in.*, 28/3 in.^. 



211. 
212. 



12 '12{bh-bji^) 



213. 


"^{R'-n,- 


. + ,.2 
2 " . 


• 


214. 


l^r%, rV3. 






215. 


/iV3, /r/3. 






216. 


7rrV2, r-/2. 






217. 


5/4 7rr\ 5/4 r\ 






219. 


1948 in.% 54.11 


in.^ 




220. 


A=:524.25sq. ft 


., 58,819 ft.*, 


, 112.19 ft.2. 


221. 


ia\ la\ 







Answeks 317 

222. l/12ah]i(a'~ + h^), 1/12 (a' + h'^). 

223. 27r'i-R{R' + ir), (R^ + ^r'); 27r'r'R{R'/2 + ir'), 

(7^72 +|r2). 

224. ^(a^ + 4/.^), ^(a^ + W); ^(a^ + h^/4), 

225. Trays. 

226. 257raV192. 

227. 4/21 Trh^h, 5/^714. 

228. -w-^^ -W^^ 

229. TVa6c(Z^^ + c^), J(^' + c')- 

230. Tra^e, a73. 

231. 180 in.^ 

232. 432 7rin.4. 

233. 0.675 lb.-ft.2 

234. 11.72 Ib.-ft.^ 

235. 17.4 Ib/in^ 

236. 374.5 lbs. 

237. 14,976 lbs. 

238. Midway. 

239. wrVi{2 + 7r/2) lbs. 

240. 468,750 lbs. 

241. 5000 lbs. 

242. 10,056 tons. 

243. 57rzm721bs. 

244. 61.1 ft. 

245. 7.74'. 

246. 4670 lbs. 

247. 5280 lbs. 

248. 2880 ft. 

249. 2275 lbs. 

250. 2h/3, 



318 Elementary Mechanics 

251. (a) /^A (b) 3/^/4. 

252. 3r/8 from vertical and 37rr/16 deep. 

253. 37rh/16. 

254. (a) 4h/7, (b) 5h/7. 

255. /i(a^ + 36)/2(a + 2&). 

256. Each top hinge 937.5 lbs., each bottom hinge 4687.5 lbs. 

257. 2.03 diameters. 

258. 1031 lbs. per linear ft., at 2.82 ft. above the sill. 

259. 12 ft. 

260. 844 lbs. at base. 

261. 2.95 ft. 

262. 2.4 ft. 

263. 18.6 ft. 

264. 9.78 ft. 

265. 28.8?^ lbs., 2.61 ft. from A and .787 ft. from AB. 

266. irr^w lbs., at depth of 5r/4 ft. 

267. 2437.5 lbs., at depth of 1.95 ft. 

268. 57r?^w;/4 Ib.-ft. 

269. 32,832 lbs., at depth of 11.21 ft. 

270. 900 lbs. 

271. 2667 lbs. per sq. in. 

272. 2:1. 



273. 


19.76 ft. 




274. 


'(ah/GVa^ + b^ 




275. 


5600 lbs. 




276. 


72tu lbs., 4 ft. below the surface 


277. 


23.6 ft. 




278. 


(a) 8.7ft., (b) 


8 . 4 f t. 


279. 


955.7 lbs. 




280. 


(a) 6w lbs., (b) 


3w lbs. 


281. 


1.69 ft. 




282. 


122,061.5 lbs. 




283. 


5.63 ft. 





Answeks 319 

284. 29,600 lbs., 35.4" below surface, and 35.1" from left end. 

285. r. 

286. 10,186 lbs. per sq. in. 

287. 4096 lbs. per sq. in. 

288. 6250 lbs. per sq. in. 

289. i" by i" in cross section. 

290. 0.17''. 

291. 79 ft. 61 in. 

292. 17,670 lbs. 

293. 17,630 ft. 

294. 14,710,000 lbs. per sq. in. 

295. 4.9, 14.6, 9.6. 

296. 49.6, 58,900 lbs. 

297. 1042 lbs. per sq. in., 48. 

298. 6875 lbs. per sq. in., 5.8. 

299. Fail by rupture of plate. 

300. 60,^^. 

301. Trqd- = 2{p — d)if where / = tensile stress. 

302. 3,394 lbs. per sq. in. 

303. 2.16". 

304. 2782.6 lbs. per sq. in., 9050.5 lbs. per sq. in., 72,^^. 

305. 0.39". 

306. 200 lbs., 831 ibs. 

307. 581 lbs., -419 lbs., -959 lbs. 

309. 160 lbs. 4' from left end, -340 lbs. 4' from right end. 

310. 4500 lbs., 500 lbs., -6500 lbs. 

311. 3.15 tons. 

312. 120 lbs. 

313. At a: = 6, 165 lbs., 2490 Ib.-ft.; at .r=10, 165 lbs., 3150 
Ib.-ft.; at a- =15, -285 lbs., 2175 Ib.-ft. 

314. 7600 Ib.-ft., 7600 Ib.-ft., 3120 Ib.-ft. 

315. At x = l^, 860 lbs., 12,600 Ib.-ft.; at .t = 19, -860 lbs., 
12,600 Ib.-ft.; at x = '2Q, -1420 lbs., 5340 Ib.-ft. 



320 Elementaky Mechanics 

319. 14f ton-ft. 

320. 1500 1b.-ft. 

321. Symmetrical; one wheel over a support. 

322. 26t ton-ft. 

323. 5.6 times as strong. 

324. Breadth (^/V3, height UI^Jl>, 64: DVItt. 

325. 520 lbs. 



327. (r,2 + r,^):r,Vr,^-r,^ (V-r/):r,\ 

328. 8. 

330. 8.64". 

331. 574.4 lbs. 

333. 3" X 10" or 2" x 12". 

335. 8.3, 5.6. 

336. 12 in. deep. 35 lbs. per ft. (steady load). 

337. 9 in. deep, 30 lbs, per ft. stationary. 12 in. deep, 40 lbs. 
per ft. rolling. 

341. -380 lbs. 

342. [)V^\. 

343. hollow/solid = 3.64. 

344. 782.15 Ib.-ins. 

345. 6484 lbs. per sq. in. 

346. 19.98 in. ' 

347. 12,011,000 lbs. per sq. in. 

348. 10.6" X 14.1". 

349. 30 X 10« lbs. per sq. in. 

350. 298.2 tons. 

351. 1.47". 

352. 2100 ft. 

353. 2.22", ^^. 

354. 2| tons per sq. in. 

355. 3f tons per sq. in. 



Answeks 321 

356. 792 lbs. 

357. 24 tons per sq. in. 

358. 239 ft. 2 in. 

359. Depth 17.32'^ breadth 10", 

360. 11,230 lbs. per sq. in. 

361. 117 lbs. 

362. -320 lbs. 

363. 1200 lbs. 

364. 2080 lbs., 12 ft. from left end. 

365. ± 144 lbs. at the center. 

367. 2400 Ib.-ft., 3600 Ib.-ft. 

368. 2560 Ib.-ft., 1120 Ib.-ft. 

369. -7200 Ib.-ft, -800 Ib.-ft. 

370. 1.82". 

371. 13.36 tons per sq. in. 

372. 5260 Ib.-ft. 

373. -466f lbs., 10.81 ft. 

374. 14.70". 

375. 55 ft. 10 in. 

376. 22 m/h. 

377. 81 f/s, 5t\ m/h. 

378. 35 m/h. 

379. 1666.6 cm/min. 

380. 8.046 kilo/hr. 

381. 3an/88^ m/h. 

382. 85 f/s, 40 f/s, 5 f/s. 

383. 0, -57r/2, -57r, -57r/2, 0, Stt/^, Stt, 57r/2, 0. 

384. 5/108 m/m^. 

385. 11 t/s\ 

386. 2.93 i/s\ 

387. 0.22 f/s\ 

388. a + 2bt + 3ct%2h + 6ct 
21 



322 Elementary Mechanics 

389. -ha sin (h + M), -Fa cos (b + kt), 

391. -2 f/s, -0.5 f/s, 0.4 t/s^, 0.025 f/s^ 

392. 2V(5-5)(s-3), 4(4-s). 

393. 40 sees. 

394. 18f/s^ 

395. 50 see., 25 metres. 

396. 5sec., 12ift. 

397. 16 f/s^ 30 f/s. 

OQQ 1 V"2-l V3-V2 

398. i sec., -^-^ — sec, ~ — — -^ — sec. 

o o 

399. 50 m/h. 

400. 24 f/s. 

401. 9 f/s. 

402. 6.25 f/s. 

403. A,0;B, 9/2 of its former velocity. 

404. 1. 

405. 9i f/s, 18i f/s. 

406. Vi. 

408. 1;= — ^cs + /^e-^^*(acos /i^-& sin M), 

409. 25 ft., i sec, 2J sec. ■ 

410. 545 cm/sec, 4/9 sec. 

411. 10.2 sec 

412. 96 f/s, 0. 

413. 20 X 10^ dynes. 

414. 49.05 kg. 

415. 2'" 56^ 

416. 110 lbs. 

417. 25:4. 

418. 6i : 1, 5/56 i/s\ 

419. 133Jf/s. 

420. 5,0625 X 10^ dynes. 



I 



Answers 323 



421. 154 lbs., 70 lbs. 

422. 20571 ft. 

423. 200 ft., 5 sec. 

424. 30°. 

425. ^(sin a — ja cos a), V2^?(sin a — /x cos a) 

427. ^/5, 7i lbs. 

428. m/2. 

429. 2 lbs. 

430. 5oz. 

431. 571 ft., lOf lbs. 

432. 24 lb. 10 oz. 

433. fi=0.3. 

434. s=a sm lc(t — tQ), 

435. Period = 27r/k. 

436. ha. 

438. 7r/2, 3, TT, 12. 

439. acoslctQ, —asinhtQ. 

440. 2 in., 0.453 sec. 

442. 42 min. 20 sees., 4.93 mi/sec. 

444. s = l/klog{l + ]vv,t). 

445. No. 

446. 2 f/s. 

447. {m^ — m2Yg/{m^-\-m^). 

448. 29 ft. 9 in. nearly. 

449. 5|o tons. 

450. 1 mi. 1408 yds. 

451. 784 ft. 

452. 3isec. 

453. Distance 6, force ^tt^ ( 7 - 25) . 

454. f±5/.,|±-:VlW. 



324 Elementaey Mechanics 

455. (5/9)*?;. 

456. 19:13. 

457. 2i lbs., 3i lbs., g/6. 

458. {m — m' sin a) g/{m-\-m'), 

459. (a) 500 ft.-lbs., (b) no. 

460. 65.82 ft.-lbs. 

461. 24,062 ft.-lbs. 

462. 352,000 ft.-lbs. 

463. 112,500 ft.-lbs. 

464. 8,906,250 ft.-lbs. 

465. 1875 ft.-lbs. 

466. 323.5 ft.-lbs. by P, 250 ft.-lbs. against R, 500 ft.-lbs. 
against gravity. 

467. 47,920 ft.-lbs. 

468. 68,760 ft.-lbs. 

469. 52 ft.-lbs., 130 ft.-lbs. 

470. (1) 48.8 ft.-lbs., (2) 195 ft. lbs. 

471. Wifxh + h) ft. 

472. 5808 ft.-lbs. 

473. 300 ft.-lbs. 

474. 338 ft.-lbs. 

475. 1.49 X 10^ ft.-lbs. 

476. SW ft.-lbs. 

477. 42,500 ft.-lbs. . ' ' 

478. 105 ft.-lbs. 

479. 500,000 ft.-lbs. 

480. 93.3 ft.-lbs. 

481. 14,242 ft.-lbs. 

482. 4.22 ft.-lbs. 

483. 26,400 lbs. 

484. 426f H. P. 

485. 508.8 H. P. 

486. 3.584 H. P. 



An'sweks 



S25 



487. 
488. 
489. 
490. 
491. 
492. 
493. 
494. 
495. 
496. 
497. 
498. 
499. 
500. 
501. 
502. 
503. 
504. 
505. 
506. 
507. 
508. 
509. 
510. 
511. 
512. 
513. 
514. 

515. 

517. 
518. 
519. 



160 H. P. 
409 H. P. 

0.20 f/s-. 

(a) 4,400,000 ft.-lbs., (b) 30 H. P. 

(a) 5.05 X 10^*^ ft.-lbs., (b) 7.65x10^ H. P 

2"^ 17.5^ 1512.5 ft. 

4.97 ft. 

300.6 ft. 

375.7 ft. 
174.5 ft. 

448.8 ft. 
742,500 ft.-lbs. 

11,778 ft.-lbs. 
1000 lbs. 
250,000 ft.-lbs. 
24,375 ft.-lbs. 
320 H. P. 
11.95 H. P. 
380 H. P. 
12 H. P. 
742.4 H. P. 
60.7 H. P. 
63.1 H. P. 
264 H. P. 
/x=.163. 



ft.-lbs. 



JL 
2- 

20.4. 
.293 

a 
w ' a. 

1V209 f/s, 32/9 f/s^. 

i;^ = 16/V~37 f/s, tv=60V37 f/s. 



326 Elementary Mechanics 

520. tanT = 2-i. 

521. Vy=-Q>V^f/s, ay^-162 i/s-, t = i sec. 

522. aa- = 4.T, ay^iy. 

524. v = 4:, at = 0, a„ = 8V2. 

525. at= -V2 i/s% an=V2 i/s\ 

526. 67r/5r/s-, ^ = 5sec. 

527. GOOtt r/m~, 6 = 75/2^- rads. 

528. -IOOtt r/nr, SOOtt rads. 

529. (a) 402, (b) 25.1 sec. 

530. —alh- cos ht, —alv- sin M, 0, al:^. 

532. ^,. = 4 f/s, Vy = 4.V3 f/s. 

533. 7.2 f/s^ 

534. 58f f/s, a = 491.7 f/s-. 

535. 1536.5 f/s, 0.111 f/s-. 

536. 27r/5 r/s^ 157r f/s, 47.57r f/s^ 

537. (a) -67r/5r/s% (b) 40f, (c) 5887r= f/s^, 

539. 2771.2 ft. 

540. a: = 886.8 ft., y = 256 ft. 

541. 45°. 

542. V = 12 V2i/s, 340 ft. 

543. 150 ft. 

544. 288 ft., 3V2 sec. 

545. 160//S, tan-^ (--i/3). 

546. 333i f/s. 

547. 6250 ft. 

548. 300 ft. nearly, t = l sec. 

549. 37,8121 ft. 

550. 5135 f/s, 51° 38'. 

551. 8V3ft. 

552. 52.7 lbs. 

553. 1 3/11 E. P. S. 

554. 0.4949 ft. 



555. 


1 hr. 25 mill. 


556. 


cos-^ 0.6079 = 5-r 3-1'. 


557. 


1.17 ft., 205.6 lbs. 


558. 


27r\/¥/g sec, ^l lbs. 


559. 


8° 36'. 


560. 


2°, 1.95 in- 


563. 


3.39 ill. 


564. 


23.39 tons. 


565. 


12 rad/sec. 


566. 


7rr*w-w/4:g. 


567. 


••-M¥-'-)- 


568. 


39.15 in. 


569. 


32.13 f/s. 


570. 


59.33. 


571. 


117^ ft. 


572. 


.g,: ^, = 125: 126. 


573. 


12V2f/s. 


574. 


4V2fAs8Vl0f/s. 


575. 


22i lbs. 


576. 


16V3 f/s. 


577. 


V/(cos^-cos^o)- 


578. 


53° 8'. 



327 



581. 1.4616/- from the lowest point of the circle. 

582. x/a = y/l), aj = Jc-x, ay = I'-ij. 

584. 324/.r^ -(16/7/)^ 

585. Vj. = ]iy/p (p is the parameter of the parabola), a^ — l'-fp^ 

587. 7^ lbs. 

588. (a) 1.5 sec, (b) 25.1 from building, (c) 59.7 f/s at 
16.5° to the vertical. 

589. 137.5 ft. from the vertical starting point, 6^ sec, 201 f/s 
at Q)^° to vertical. 



328 Elementaey Mechanics 

590. a,= -^VVTo;a,=:6VV^90. 

592. (a) 0.022 r/s, (b) 15.7 f/s, (c) 7.85 f/s. 

593. 196 ft. 

594. Arc tan f or arc tan 14/3. 

595. ^-^I^i.rdeidi. 

596. 46Tf ^ ft.-lbs. 

597. 6.52 in. 

599. 6.16 lbs. per ton. 

600. 302.5 ft. 

601. 41 m/h (nearly), 

602. 1053 H. P. 

603. 19.2 m/h. 

604. 90^. 

605. 7.23 in. 

606. 19,950 ft.-lb., 241.8 H. P. 

607. 7.37 sees. 

608. 5/3 lbs. 

609. 91 f/s^ 18f r/s^ 14| lbs. 

610. y8=0.335r/s^ T=49.91 lbs., s = 0.698 ft., a=:0.056 f/s' 

611. 23 rev. 

612. 7rVlla/6^. 

613. 2.3338 ft. 

614. 92.6 beats per min. 

615. h^r. 

616. 0.463 sec, 0.7 ft., 0.2 ft. below center of sphere. 

617. 0.866 sec, 2.44 ft. 

618. 125 (nearly). 

619. 3300 lb.-ft.2. 

620. 48.7 ft.-lbs. 

621. 44.3 revolutions. 

622. 20,600^2 ft.-lbs., 10.5 revolutions. 

623. 2 xl02» ft.-lbs. (approx.). 



624. 


208 tons. 




625. 


3367.3 H. P. 




626. 


14.57 in. 




627. 


External diam, 


. 15.09 in., i: 


lbs. 






628. 


4itons, 98f^. 




629. 


96 f/s, 432 ft., 


19.59 f/s. 


630. 


280 ft. 





329 



internal diam. 8.49 in., 8893.5 



631. Sphere, disk, hoop. 

634. (a) 11.7 f/s downward, (b) 5.9 r/s, (c)-W~r/s^ W- y^2^ 
-W- f/s', (d) 1.^03 sec. 

635. 1,391,500 ft.-lbs. 

636. 26.89 ft.-tons. 

637. 2611 f/s, 6596 ft.-lbs., 30.37 lbs. 

638. 8 f/s. 

639. 250 ft.-lbs. 

640. 139.57 tons, 271.38 ton-ft. 

641. 7|lbs. 

642. 1000 ft.-lbs. 



644. 18,497.03 ft.-tons. 

645. 37.5 revolutions. 

646. 8.3 lbs. per ton, 81 m/h. 

647. 1/36 lb.-ft.2. 

648. 328.17 ft.-tons, 928 revolutions. 

649. 8 tons. 

650. 927,778 ft.-lbs. 

651. 8.04 ft. 

652. 7230 ft.-lbs. 



